Stochastic Differential Equation vs Ordinary Differential Equation

An ordinary differential equation and a starting value determine exactly one path. A stochastic differential equation and the same starting value determine a whole distribution of paths instead. The two equations differ in what gets determined, not in how hard either is to handle, and nearly every other difference between them follows from that one.

One picture is worth carrying before any notation arrives. A good kitchen scale, put under the same bag of rice ten times, reads 250 grams ten times. A cheap one reads 248, then 252, then 249, then 253. Both scales are describing the same bag. The first gives a number and stands behind it. The second gives a spread, and the middle of that spread may be exactly 250. The difference between the two equations in this guide is the difference between those two scales: not a difference in accuracy, but a difference in the kind of thing that comes out.

The analogy carries one idea and then stops, so it is worth naming what it does not carry. The randomness in a stochastic differential equation is not measurement error sitting on top of a true path. There is no true path underneath it that better instruments would reveal. The spread is the answer, not noise around the answer, and that is exactly what makes the comparison worth setting out in full.

Each equation is defined in full and in its own right below, from scratch, before either is set against the other. Nothing is contrasted until both pictures are complete.

Every worked number below belongs to one invented quantity, the standard process, written S with a time subscript. The standard process starts at Rs 100/-, grows at 8 per cent a year, carries a volatility of 20 per cent a year, and is followed to a horizon of one year. Those four numbers generate every figure that follows.

What is an ordinary differential equation, and what is it built from?

An ordinary differential equationAn equation of change with no random term in it, which together with a starting value determines exactly one path. is a rule that ties the rate at which a quantity is changing to the quantity itself, and possibly to the time. The rule never names the quantity. The rule gives the speed of the quantity, given where the quantity currently is. Saying that much is all an equation of this kind ever does.

On its own such a rule is not enough to pin anything down. A rule saying that the growth rate is 8 per cent a year is equally satisfied by a path starting at Rs 100/-, a path starting at Rs 1/-, and a path starting at Rs 4,000/-. The second ingredient is the initial conditionThe value the quantity takes at the start, which together with the equation fixes which one path is the answer., the value the quantity takes at the beginning. Rule plus starting value is the whole recipe, and out of it comes exactly one path.

Formula block one, the ordinary differential equation and the single path it determines

$$ \frac{dS_t}{dt} \;=\; \mu\, S_t, \qquad S_0 = 100, \qquad\Longrightarrow\qquad S_t \;=\; S_0\,e^{\mu t} $$

$S_t$	the standard process, the invented quantity being modelled, at time $t$
$dS_t/dt$	the rate at which that quantity is changing at time $t$, in rupees a year
$\mu$	the growth rate, here 0.08, meaning 8 per cent a year
$S_0$	the starting value, here Rs 100/- exactly
$t$	elapsed time in years, running from 0 to the horizon $T=1$

What it says in wordsAt every instant the quantity is growing at a fixed multiple of whatever it currently is, and once the starting value is also stated, there is exactly one path in the world that obeys both statements at once. That path is the starting value multiplied by a growing exponential factor.

The random term quietly destroys two properties of that answer, so both deserve naming while they are still there. The first is that the answer is deterministicHaving exactly one outcome, with nothing at all left to chance once the inputs are fixed.. Asked where the quantity is at six months, the equation gives one answer, Rs 104.081077/-, and there is no second one. Asked again tomorrow, with the same equation and the same starting value, it gives Rs 104.081077/- again. The equation has no memory of having been asked and no scope to answer differently.

The second property is that this particular equation happens to have a closed form solutionA formula that gives the answer directly, without any stepping or approximation. Some equations of both kinds have one and most do not.. Having a formula is a convenience of this example rather than a feature of the type. Most ordinary differential equations that people actually meet have no formula and are stepped forward on a machine instead. The formula also lets every number that follows be checked by hand.

The ordinary equation at 8 per cent a year fixes a single path from Rs 100/- through Rs 104.081077/- at six months to Rs 108.328707/- at the horizon.

Try it out

Given the rule that the growth rate is 8 per cent a year, and nothing else. How many paths satisfy it?

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What is a stochastic differential equation, and what is it built from?

A stochastic differential equation is built from three ingredients rather than two, and the third one is what changes the character of the whole object. The first ingredient is a drift rule, saying how the quantity tends to move over a short stretch of time. The second is a diffusion rule, saying how hard it is pushed around over that same stretch. The third is the source of the pushing. Here that source is Brownian motion, written W with a time subscript. Add the starting value and the recipe is complete.

Written down, it looks like an ordinary differential equation with one extra term bolted on. The appearance is misleading in a specific and important way, and the way is worth spelling out. A Brownian path has no rate of change at any point, so the letter d in front of W does not mean a rate of change. The differential form is shorthand, and the thing it is shorthand for is an integral equation.

Formula block two, the stochastic differential equation and the integral equation it stands for

$$ dS_t \;=\; \mu\,S_t\,dt \;+\; \sigma\,S_t\,dW_t $$ $$ S_t \;=\; S_0 \;+\; \int_0^t \mu\,S_u\,du \;+\; \int_0^t \sigma\,S_u\,dW_u $$

$S_t$	the standard process, the invented quantity being modelled, at time $t$
$\mu$	the drift, here 0.08, meaning 8 per cent a year
$\sigma$	the volatility, here 0.20, meaning 20 per cent a year
$W_t$	standard Brownian motion under the physical measure P, the source of the randomness
$dW_t$	the increment of that Brownian motion, which is not a rate and cannot be divided by $dt$
$u$	the integration variable, running over the stretch from 0 to $t$
$S_0$	the starting value, here Rs 100/- exactly

What it says in wordsOver any stretch of time the quantity moves by two contributions added together: a steady pull proportional to its own level and to the time elapsed, and a push proportional to its own level and to the movement of the driving Brownian motion over that stretch. The first line is shorthand for the second, and the second is the statement that actually means something, because it asks only for integrals and never for a slope.

Notice what has and has not changed. The drift term is doing exactly the job the whole ordinary equation was doing. The diffusion term is new, and it is attached to a driving path that wanders in a way no ordinary equation ever has to accommodate. Because the driving path is different on every occasion the equation is realised, the object the equation determines is no longer a path. The object is a distribution of pathsThe whole set of paths the equation permits, each carrying the weight the probabilities give it, treated as one object., meaning the whole set of them with the weights probability puts on them.

Both equations need a rule and a starting value, and the stochastic one adds a diffusion rule and a driving Brownian motion on top.

One member of that set is worth drawing out. The locked path is a twelve step path published once and reused wherever the standard process appears, so every treatment draws the same one and a reader can check a picture against a table. The locked path runs from Rs 100/- to a high of Rs 111.08/- at month six, down to a low of Rs 93.74/- at month nine, and finishes at Rs 106.18/-. The locked path is not the answer to the stochastic equation. The locked path is one of the objects the answer contains, and treating a single drawn path as the answer is the standard mistake.

The locked path shows what one realisation of the stochastic equation looks like, moving between Rs 93.74/- and Rs 111.08/- on its way to Rs 106.18/-.

Try it out

Someone presents the locked path and calls it the solution of the stochastic equation. What is wrong with that?

What does each one determine from the same Rs 100/-?

Both objects are now on the table, so the contrast can start. Give each equation the identical starting value of Rs 100/-, the identical growth or drift of 8 per cent a year, and the identical one year horizon. The only thing the second one has that the first does not is a volatility of 20 per cent. Now ask each of them the same question: where is the quantity at the horizon?

The ordinary equation answers Rs 108.328707/-. One number is the entire answer. There is nothing else in it, no second number, no range, no qualification. The stochastic equation cannot answer with a number at all, since the question does not have a number as its answer. The answer is a shape, and a shape has to be described by naming several things about it at once.

Formula block three, the solution of the stochastic equation

$$ S_t \;=\; S_0\,\exp\!\Bigl(\bigl(\mu - \tfrac{1}{2}\sigma^{2}\bigr)t \;+\; \sigma W_t\Bigr) $$

$S_t$	the standard process at time $t$, now a random quantity rather than a number
$S_0$	the starting value, Rs 100/- exactly
$\mu$	the drift, 0.08
$\sigma$	the volatility, 0.20, so $\sigma^{2}$ is 0.04 and half of it is 0.02
$W_t$	standard Brownian motion under the physical measure P, which supplies the randomness
$\exp$	the exponential function, so the whole right hand side is always positive

What it says in wordsThe quantity at any time is its starting value multiplied by an exponential whose exponent has a steady piece and a random piece. The steady piece is not the drift but the drift less half the variance rate, and the random piece is the volatility multiplied by wherever the driving Brownian motion happens to be. Because the exponent is normally distributed, the quantity itself is always positive and its spread leans to the right.

The right hand side is always positive and its spread is lognormalDescribing a quantity whose logarithm is normally distributed, so it can never go below zero and its spread leans to the right.. That single structural fact sits behind everything that follows. Three numbers describe that shape, and they are three different numbers rather than one number described three ways. The average outcome is Rs 108.328707/-. The middle outcome, the one with half the weight above it and half below, is Rs 106.183655/-. The most likely outcome, the peak of the distribution, is Rs 102.020134/-. All three come from the same distribution and none is more the answer than the others.

Formula block four, the three centres of the distribution at the horizon

$$ \mathbb{E}[S_T] = S_0e^{\mu T}, \qquad \operatorname{med}(S_T) = S_0e^{(\mu-\frac12\sigma^{2})T}, \qquad \operatorname{mode}(S_T) = S_0e^{(\mu-\frac32\sigma^{2})T} $$

$\mathbb{E}[S_T]$	the average of the quantity at the horizon, Rs 108.328707/-
$\operatorname{med}$	the middle outcome, with half the weight either side, Rs 106.183655/-
$\operatorname{mode}$	the most likely outcome, the peak of the distribution, Rs 102.020134/-
$T$	the horizon, one year
$\sigma^{2}$	the variance rate, 0.04, so half of it is 0.02 and one and a half times it is 0.06

What it says in wordsAll three centres are the starting value multiplied by an exponential, and they differ only in how much of the variance rate is subtracted from the drift before the exponential is taken. Subtracting nothing gives the average. Subtracting half the variance rate gives the middle. Subtracting one and a half times it gives the peak. Because the variance rate is positive, they come out in that order every time.

Now put the two answers beside each other and read the first line of each. The average of the stochastic answer is Rs 108.328707/-, and the ordinary answer is Rs 108.328707/-, and those are the same number to the last decimal place shown. That is not a coincidence and it is not an approximation that happens to be good at these parameters. The equality is exact, and the reason is worth its own formula block.

Formula block five, why the ordinary equation returns the average exactly

$$ \mathbb{E}\!\left[\int_0^t \sigma S_u\,dW_u\right] = 0 \qquad\Longrightarrow\qquad \frac{d}{dt}\,\mathbb{E}[S_t] \;=\; \mu\,\mathbb{E}[S_t], \qquad \mathbb{E}[S_0]=100 $$

$\mathbb{E}[\,\cdot\,]$	the average taken over the whole distribution of paths
$\int_0^t \sigma S_u\,dW_u$	the diffusion term of the stochastic equation, gathered up to time $t$
$\mu$	the drift, 0.08, the same number the ordinary equation uses as its growth rate
$\mathbb{E}[S_t]$	the average of the quantity at time $t$, which is what the right hand equation is about

What it says in wordsTake the average of every term in the stochastic equation. The diffusion term averages to nothing, because the driving Brownian motion is as likely to push one way as the other and the integral inherits that. What is left is an equation in which the average changes at a fixed multiple of the average, which is the ordinary equation with the average written in place of the level. So the ordinary equation is not an approximation to the stochastic one. It is exactly the equation the average of the stochastic answer satisfies.

The exact equality reorganises the whole comparison. The ordinary equation is not a cruder tool that gets close. The ordinary equation is a precisely correct tool answering a narrower question. It returns the average, to the last decimal, and it is silent about everything else: the middle, the peak, the spread, the chance of finishing below the starting value, all of it. Everything else is what stochastic calculus exists for.

The ordinary equation returns the average of the stochastic answer exactly and reports nothing about the middle, the peak or the spread.

Try it out

The ordinary answer is Rs 108.328707/- and the stochastic average is Rs 108.328707/-. Coincidence?

Try it out

An ordinary equation together with a starting value determines how many paths?

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How does the ordinary path sit inside the stochastic answer?

Drawn together, the relationship is easy to see and easy to misread. The distribution at each time can be shown as a fan of bands, with the widest pair of edges holding the middle ninety per cent of outcomes and the inner pair holding the middle half. The starting value is known, so at the start the fan is a point. As time runs the fan opens out, and by the horizon its ninetieth percentile edges sit at Rs 76.416563/- below and Rs 147.546136/- above.

The ordinary path runs right through that fan as the mean pathThe path traced by the average of the distribution at each time, which here is exactly the path the ordinary equation determines.. At every single time, not just at the horizon, the ordinary path equals the average of the stochastic distribution at that time. One line is drawn, and that one line does two jobs. The line called the ordinary answer and the line called the average of the stochastic answer are the same line.

Here is where the misreading starts. Being the average does not make it the middle. The middle line, drawn dashed just below it, is the path of the quantileA cut point of a distribution. The fiftieth quantile has half the weight below it, the fifth has five per cent below it, and so on. that has half the weight either side, and by the horizon it has fallen Rs 2.145052/- behind. On the scale of a fan running from Rs 76/- to Rs 148/- that gap is barely a hair. A magnified strip sits beside the figure for exactly that reason. Read on the wide scale the two lines look identical. Read on the magnified scale they are plainly not.

The ordinary path is the average of the stochastic distribution at every time, and the middle outcome falls Rs 2.145052/- behind it by the horizon.

The ladder of three centres makes the same point without the fan getting in the way. Laid out on a rupee scale from Rs 100/- to Rs 110/-, the peak of the distribution sits at Rs 102.020134/-, the middle at Rs 106.183655/- and the average at Rs 108.328707/-. The ordinary equation reports the rightmost of the three and does not know the other two exist.

The most likely, the middle and the average outcomes are three different numbers, and only the last of them is what the ordinary equation returns.

Try it out

The volatility is about to rise. Before the control moves: does the ordinary path move?

Play with it

Move the volatility and watch what refuses to move

One control, the volatility, from 0 to 40 per cent a year. The drift stays at 8 per cent and the horizon stays at one year. The fan of outcomes redraws at every setting. No volatility appears anywhere in the equation that produced the solid line, so the solid line stands still. The strip on the right is a magnified view of the horizon, where the middle outcome walks away from a line that is standing still.

Jump to a named setting

Volatility

20 per cent

Ordinary answer

Rs 108.328707/-

Middle outcome

Rs 106.183655/-

Share reaching the ordinary answer

46.0172 per cent

At a volatility of 20 per cent a year the ordinary equation answers Rs 108.328707/- at the horizon, the middle outcome of the stochastic answer sits Rs 2.145052/- below it at Rs 106.183655/-, and 46.0172 per cent of outcomes reach the ordinary answer.

Volatility	Ordinary answer	Middle outcome	Share reaching it
0 per cent	Rs 108.328707/-	Rs 108.328707/-	50.0000 per cent
20 per cent, the worked setting	Rs 108.328707/-	Rs 106.183655/-	46.0172 per cent
30 per cent	Rs 108.328707/-	Rs 103.561971/-	44.0382 per cent
40 per cent	Rs 108.328707/-	Rs 100.000000/-	42.0740 per cent

Educational illustration. Every reading is computed from the formula and never sampled, so the default of 20 per cent reproduces the worked instance exactly on every reload. The ordinary answer reads Rs 108.328707/- in all four rows because the ordinary equation contains no volatility term at all, and that immobility is the finding rather than a fault in the drawing. At 40 per cent the middle outcome lands on exactly Rs 100.000000/-, so half of outcomes finish at or below the starting value while the ordinary equation still answers Rs 108.328707/-. The drift stays at 8 per cent a year and the horizon at one year throughout.

What does it mean to solve each of them?

Solving the ordinary equation means producing the path. Either a formula is found for it, as was done above, or it is stepped forward on a machine to produce a table of values. Either way the finished object is a list of numbers, one per time, and it can be handed to somebody who reads a value off it. Consider a lift that only reports which floor it is on. Solving the ordinary equation supplies the floor at every moment of the day.

Solving the stochastic equation means something else, and the difference is not that it is harder. The finished object is a different kind of thing. What is wanted is a process, meaning a rule that assigns a path to each realisation of the driving Brownian motion, together with the distribution that rule induces at each time. There is no single list of values, so no list can be handed over. The shape at the horizon can be handed over instead, along with a rule for computing the chance of any statement about the path being true.

What counts as having solved a stochastic differential equation is set out under strong and weak solutions. More than one answer is sensible there, which a reader arriving from the ordinary side rarely expects. Here the two activities differ in kind, not merely in difficulty.

The question asked	What the ordinary equation returns	What the stochastic equation returns
Where is it at six months?	Rs 104.081077/-	A shape whose average is Rs 104.081077/- and whose middle is Rs 103.045453/-
Where is it at one year?	Rs 108.328707/-	A shape whose average is Rs 108.328707/-, middle Rs 106.183655/-, peak Rs 102.020134/-
What is the chance it finishes below Rs 100/-?	No statement available	A computable number
What does the whole answer look like?	One path	A distribution of paths

Try it out

Solving an ordinary equation gives a path. What does solving a stochastic one give?

Which techniques carry across, and which have no counterpart?

A reader arriving from the ordinary side has a toolbox, and the useful question is which of those tools still work. The answer sorts into three groups rather than two, and the third group is the one that catches people out because there is nothing to fix in it.

The tools that carry across untouched are the structural ones. Linearity carries: constants factor out of both terms, and if two rules are added the answers add. The idea that a starting value pins down the answer carries: it just pins down a distribution instead of a path. Superposition for linear equations carries in the same modified sense. Anything in the ordinary toolbox that was about the shape of the equation rather than about the smoothness of the path still applies.

The tools that carry across with one modification are the calculus ones. The chain rule still applies to a function of the process, but it gains one extra term, and that extra term is the whole reason the exponent in the solution carries a subtraction of half the variance rate rather than nothing. The extra term comes from Ito's lemma, which is set out under Ito calculus and used here rather than rebuilt.

The tools with no counterpart are the ones that quietly assumed the path had a slope. Separation of variablesThe standard ordinary technique of gathering one variable on each side of the equation and integrating both sides. Separation has no counterpart against a path with no slope. is the headline case: the quantity gathers on one side and the time on the other, both sides are integrated, and the work is finished. The driving term is not a function of time that can be moved to the other side, so against a stochastic equation there is nothing to separate. Dividing through by the increment to recover a rate is the second case, and it fails for the plainest possible reason: no point of the path has a rate at all. Neither of these is a rule that needs an extra term. There is no first term to correct.

Structural tools carry across untouched, the chain rule survives with one extra term, and two ordinary techniques have no counterpart whatsoever.

Try it out

Name an ordinary technique that has no counterpart against a stochastic equation.

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When is the ordinary equation the right choice?

Often, and a comparison that did not say so would be an argument dressed as a comparison. The test is a single question, and it is a question about what is being asked rather than about the mathematics. Is the spread part of what needs to be known?

If the spread is not part of the question, the ordinary equation is the better instrument on every count. The average it returns is the average of the stochastic answer exactly, so the ordinary equation answers the question actually asked. Setting it up, stating it and computing it cost a fraction as much. The ordinary equation also needs one parameter fewer, and the parameter it drops is the volatility, the hardest of the four to pin down. Reaching for the stochastic equation when only the average is wanted buys nothing and pays for it in every direction.

If the spread is part of the question, no amount of care with the ordinary equation will recover it. The simpler tool does not give a rough version of the right answer. The simpler tool gives one exact number out of a shape with many features, and the features it omits cannot be reconstructed from the number it supplies. A middle cannot be inferred from an average, nor a chance of falling short from either.

When the spread is genuinely irrelevant the ordinary equation answers correctly and far more cheaply, and only otherwise is the extra machinery earned.

Try it out

The question is about the average alone and the spread is genuinely irrelevant. Which equation?

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How does somebody reading a projection use this?

Building either equation is not necessary to get value from the distinction. Three checks, all of which can be run on a printed projection with nothing else to hand, catch the great majority of what goes wrong when a single line is handed over as though it were the whole answer.

Ask which centre the single line is A projection drawn as one line is reporting one number per time, and there are at least three candidates. Ask whoever produced it whether the line is the average, the middle or the most likely outcome. If they have not thought about the question, the line is almost certainly the average, because the average is what falls out of an ordinary equation.
On the standard process those three answers are Rs 108.328707/-, Rs 106.183655/- and Rs 102.020134/-, which is a spread of Rs 6.308573/- across a projection that looked like one number.
Ask what share of outcomes reach the line If the line is the average of a right leaning distribution, the honest share is below one half and it falls as the volatility rises. A projection that cannot answer this is not wrong, but it is narrower than it looks, and the narrowness is invisible on the printed sheet.
At a volatility of 20 per cent the share is 46.0172 per cent, at 30 per cent it is 44.0382 per cent, and at 40 per cent it is 42.0740 per cent.
Ask what would change if the volatility were doubled This is the cheapest test of all, because the answer reveals which kind of equation is underneath without seeing it. If nothing on the projection would move, it came from an ordinary equation. If the middle would slide down while the headline stayed put, it came from a stochastic one.
Doubling the volatility from 20 to 40 per cent moves the middle outcome from Rs 106.183655/- to Rs 100.000000/- and leaves the average at Rs 108.328707/- exactly.

All three checks are the same question asked three ways: is this a number or is it a shape, and if it is a number, which feature of the shape is it? The question survives being asked about projections built with mathematics the reader has never seen, so it is worth carrying long after every formula above has faded.

The error that gets made, and what it costs

Treating the ordinary answer as the stochastic answer with the randomness taken out. That reading is easy and it is wrong. The ordinary answer is not a cleaned up version of anything. The ordinary answer is the average of the distribution, and on a distribution that leans to the right the average is not the typical outcome and is not the middle one.

Consider counting a queue. Knowing that the average length of a queue over the day is eleven people says nothing about how often it is empty and nothing about how often it runs out of the door, and a person planning the counter staffing around eleven is planning around a number that describes neither of the situations that actually cause trouble. The average is a real fact about the queue. The average is just not the fact that was needed.

Here the same thing happens with numbers that can be checked. A plan built on the ordinary path is a plan built on Rs 108.328707/-, and 46.0172 per cent of outcomes reach it. Fewer than half. The middle outcome is Rs 106.183655/-, a shortfall of Rs 2.145052/- against the plan, and the shortfall widens as the volatility rises: at 40 per cent the middle outcome is exactly Rs 100.000000/-, meaning half of outcomes finish at or below where the whole thing started.

The cost is a plan resting on a figure the majority of outcomes do not reach, produced by arithmetic that is entirely correct. Nothing was miscalculated. The ordinary equation was solved properly and it answered a question about the average that nobody had asked. The absence of any wrong number to find is what makes this failure hard to catch.

Fewer than half of outcomes reach the ordinary answer at any positive volatility, and the share falls further as the volatility rises.

The drift and diffusion terms are opened separately under the components of a stochastic differential equation. What counts as having solved either equation is set out under strong and weak solutions. Numerical methods for stepping either one forward are covered under discretisation. Pricing arguments are covered separately, and what any contract pays is taken as known here. No jurisdiction anywhere sets the form of a differential equation, so no rule, threshold, rate or period is engaged.

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References

Source	Document	Where
arXiv Quantitative Finance	Preprint repository for work on stochastic differential equations and their solutions	arxiv.org
Social Science Research Network	Working paper repository for the same material	ssrn.com
Hull, Shreve and Wilmott	Standard texts, consulted for structure and notation only, with no text reproduced	named in the text, no text reproduced
Kiyosi Ito	The chain rule for a function of a stochastic process, the source of the correction of half the variance rate in the solution	named in the text

The standard process and the locked path are invented.
Educational material. Not advice on any investment, tax, budget or market position.

\(S_t\)	the standard process, the invented quantity being modelled, at time \(t\)
\(dS_t/dt\)	the rate at which that quantity is changing at time \(t\), in rupees a year
\(\mu\)	the growth rate, here 0.08, meaning 8 per cent a year
\(S_0\)	the starting value, here Rs 100/- exactly
\(t\)	elapsed time in years, running from 0 to the horizon \(T=1\)

\(S_t\)	the standard process, the invented quantity being modelled, at time \(t\)
\(\mu\)	the drift, here 0.08, meaning 8 per cent a year
\(\sigma\)	the volatility, here 0.20, meaning 20 per cent a year
\(W_t\)	standard Brownian motion under the physical measure P, the source of the randomness
\(dW_t\)	the increment of that Brownian motion, which is not a rate and cannot be divided by \(dt\)
\(u\)	the integration variable, running over the stretch from 0 to \(t\)
\(S_0\)	the starting value, here Rs 100/- exactly

\(S_t\)	the standard process at time \(t\), now a random quantity rather than a number
\(S_0\)	the starting value, Rs 100/- exactly
\(\mu\)	the drift, 0.08
\(\sigma\)	the volatility, 0.20, so \(\sigma^{2}\) is 0.04 and half of it is 0.02
\(W_t\)	standard Brownian motion under the physical measure P, which supplies the randomness
\(\exp\)	the exponential function, so the whole right hand side is always positive

\(\mathbb{E}[\,\cdot\,]\)	the average taken over the whole distribution of paths
\(\int_0^t \sigma S_u\,dW_u\)	the diffusion term of the stochastic equation, gathered up to time \(t\)
\(\mu\)	the drift, 0.08, the same number the ordinary equation uses as its growth rate
\(\mathbb{E}[S_t]\)	the average of the quantity at time \(t\), which is what the right hand equation is about