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  1. 003X and Y are independent and each uniform on 0 to 1. What is the probability that X + Y is less than 1.5, and what shape is the density of X + Y?Continuous probability and distributionsCoreCitadelChicago · 2025

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    The probability is 7/8, and the density of X + Y is a triangle, a tent peaking at 1. Because X and Y are independent and uniform, every point of the unit square is equally likely, so probability is area. The line x + y = 1.5 slices off a corner triangle with legs of 0.5, area 1/8. The sum's density rises in a straight line from 0 to 1 and falls back to 0 at 2.

    Why does probability become area here?

    Throw a dart at a square board so that every point is equally likely to be hit. The chance it lands in a region is that region's share of the board. Two independent uniforms are exactly that dart: the pair (X, Y) lands evenly on the unit square, so any question about X and Y becomes a question about an area. The condition X + Y below 1.5 is everything under the line x + y = 1.5, which is the whole square except one corner.

    Probability is area: the missing corner is 1/8, so the answer is 7/8x + y = 1.51/8X + Y below 1.5area 1 - 1/8 = 7/8000.50.511XY011.521peak at 1: the tenttail beyond 1.5area 1/8Density of X + Y
    The line x + y = 1.5 removes a corner triangle of area 1/8 from the unit square, so X + Y is below 1.5 with probability 7/8, and the density of X + Y is a tent on 0 to 2 whose tail beyond 1.5 also has area 1/8.

    How do you get the shape of the sum's density?

    Slide the line x + y = s across the square and watch how long it is inside. Near s = 0 it barely clips the corner; at s = 1 it runs corner to corner, the longest it gets; past 1 it shortens again. The density of the sum at s is proportional to the length of that line inside the square, which gives a triangle rising from 0 to a peak at 1 and falling to 2. This is the convolutionThe density of a sum of independent variables, found by adding up every way the two parts can combine to the same total. of two flat densities, and the same reason two dice most often total 7.

    The relationship
    fX+Y(s)=∫01fX(x) fY(s−x) dx={s0≤s≤12−s1≤s≤2f_{X+Y}(s) = \int_0^1 f_X(x)\, f_Y(s-x)\, dx = \begin{cases} s & 0 \le s \le 1 \\ 2 - s & 1 \le s \le 2 \end{cases}
    f_{X+Y}(s)the density of the sum at the value s
    f_Y(s - x)equal to 1 when s - x lies between 0 and 1, otherwise 0
    What it says in wordsAdd up every split of s into an x and a y that both lie in 0 to 1; the count of splits rises to s = 1 and then falls.

    Check the first answer with the tent. The area beyond 1.5 is a triangle with base 0.5 and height 0.5, which is 1/8 again. Two routes that agree is the check an interviewer wants to hear before you commit. Add a third uniform and the density becomes three joined curved pieces; add many and the sum looks normal, which is the central limit theorem arriving in slow motion.

    Where candidates lose it

    Candidates reach for a double integral before drawing, set the limits wrongly, and spend two minutes on what is a one-line area argument. Draw the square first; the corner triangle is visible at a glance.

    The second loss is saying the sum of two uniforms is uniform on 0 to 2. It is not: there is only one way to get a sum near 0 and many ways to get a sum near 1, which is why the density is a tent and not a flat line.

    What the interviewer asks next

    • What is the probability that X + Y is less than 0.5?
    • What is the probability that the larger of X and Y is below 0.5, and how does the picture change?
    • What does the density of X + Y + Z look like?

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