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  1. 063X and Y are independent random variables, each uniform on 0 to 1. What is the density of X + Y, and what is the probability that X + Y is less than 1.5?Continuous and geometric probabilityWarm upCitadelChicago · 2025

    Try it first

    What is P(X + Y < 1.5)?

    Show the worked solution

    The density is a triangle, f(s) = s for s up to 1 and 2 - s from 1 to 2, and P(X + Y < 1.5) = 7/8. Convolving two flat densities gives a tent peaking at 1. The part above 1.5 is a triangle with base 0.5 and height 0.5, area 1/8. In the unit square it is the same corner: the line x + y = 1.5 cuts off a triangle with legs of 0.5.

    Why is the sum not uniform on 0 to 2?

    Roll two dice: a total of 7 can be made six ways, a total of 12 only one way. Continuous uniforms behave the same. A sum near the middle can be made from many pairs, a sum near either end from very few, so the density of the sum rises to a peak and falls again. The mechanism that builds it is {term('convolution', 'The density of a sum of independent variables: for each possible total, add up the density of every pair of values that makes it.')}: the density at s is the length of the set of x values for which both x and s - x lie between 0 and 1.

    Two flat densities convolve into a triangle; the tail above 1.5 is 1/81/8X + Y < 1.5area 7/8x + y = 1.5000.50.511XY00.511.521value of X + Ypeak 1 at sum 1rises: f(s) = stail above 1.5:(1/2)(0.5)(0.5) = 1/87/8
    In the unit square the line x + y = 1.5 cuts off a corner triangle of area 1/8, and in the triangular density of the sum the tail above 1.5 is the same 1/8, so X + Y is below 1.5 with probability 7/8.

    How does the convolution give the triangle?

    Fix a total s. You need x between 0 and 1 and also s - x between 0 and 1, so x must lie between max(0, s - 1) and min(1, s). For s below 1 that interval has length s, and for s above 1 it has length 2 - s, so the density is a tent with its peak of 1 at s = 1. Check that the area is 1: a triangle with base 2 and height 1. The mean is 1 and the variance is 1/12 + 1/12 = 1/6, both of which you can read from symmetry and independence.

    The relationship
    fX+Y(s)=∫011{0≤s−x≤1} dx={s0≤s≤12−s1≤s≤2P(X+Y<1.5)=1−12(0.5)2=78f_{X+Y}(s) = \int_0^1 \mathbf{1}\{0 \le s-x \le 1\}\,dx = \begin{cases} s & 0\le s\le 1\\ 2-s & 1\le s\le 2\end{cases} \qquad P(X+Y<1.5) = 1 - \tfrac12(0.5)^2 = \tfrac78
    f(s)the density of the sum at the value s
    the indicator1 when s - x is a valid value of Y, otherwise 0
    What it says in wordsThe density of the sum at s is how many ways of splitting s are allowed, which rises linearly to 1 and falls back.

    Why give the square picture as well?

    It is a check that costs ten seconds. Because the pair is uniform on the unit square, any probability about X + Y is an area, and the event X + Y at least 1.5 is the corner triangle above the line x + y = 1.5. Its legs run from 0.5 to 1 on each axis, so its area is 1/8 and the answer is 7/8 again. The density gives you the whole distribution; the square gives you any single probability fast. Keep both, because the next question is usually three uniforms, where the density becomes piecewise quadratic and the square becomes a cube: P(X + Y + Z < 1) = 1/6.

    Where candidates lose it

    The common slip is 3/4, from assuming a sum of uniforms is uniform. Sums are never uniform unless one of the pieces is degenerate; they pile up in the middle.

    The second loss is getting the convolution limits wrong and producing a density that does not integrate to 1. Write the two constraints on x out loud, and check the triangle's area before you use it.

    What the interviewer asks next

    • What is the density of X - Y?
    • What is P(X + Y + Z < 1) for three independent uniforms?
    • What is the expected value of max(X, Y), and of X + Y given that X + Y > 1?

    Asked at Citadel, Quant Research Interview, Chicago, 2025 (Wall Street Oasis): He was asking some questions about the probability, especially on the convolution.