Convolution and Sums of Random Variables

You know the distribution of Monday's return and the distribution of Tuesday's return, independently. What is the distribution of the two-day return, their sum? You cannot just add the two densities together — that would answer a different question ("what's the density of a return that's either Monday's or Tuesday's"). Adding two independent random variables mixes their randomness in a specific way, and the operation that computes the resulting density is called convolution.

The analogy: rolling two dice, slid past each other

Think about rolling two ordinary dice and asking for the distribution of the total. To get a total of 7, you can have (1,6), (2,5), (3,4), (4,3), (5,2), or (6,1) — six different ways the first die's value and the second die's "gap to 7" line up. To find $P(\text{total}=7)$, you check, for every possible value $k$ of the first die, whether $7-k$ is a possible value of the second die, and add up the matching probabilities. That sliding-and-matching process — hold the total fixed, slide one distribution against the other, and sum wherever they overlap — is exactly what convolution does, generalized from dice to any distribution, discrete or continuous.

Writing it down

If $X$ and $Y$ are independent with $Z = X + Y$, the distribution of $Z$ is the convolution of the distributions of $X$ and $Y$. For discrete variables,

$$P(Z=z) = \sum_{k} P(X=k) \, P(Y = z-k) .$$

In words: to land on total $z$, $X$ must land on some value $k$ and $Y$ must land on exactly $z-k$ — sum that joint probability over every possible split $k$. For continuous variables the sum becomes an integral,

$$f_Z(z) = \int_{-\infty}^{\infty} f_X(x)\, f_Y(z-x)\, dx ,$$

which says the same thing with densities: slide $Y$'s density backwards by $x$ for every possible $x$, weight by how likely $X=x$ is, and integrate. This is why convolution is sometimes described as "slide, multiply, and sum" — one density is dragged across the other and, at each position, the overlap gets weighted and accumulated.

Worked example 1: sum of two dice, computed by convolution

Let $X$ and $Y$ each be a fair die, $P(X=k)=P(Y=k)=1/6$ for $k=1,\dots,6$. Find $P(X+Y=5)$. List the splits $k$ where both $P(X=k)>0$ and $P(Y=5-k)>0$: $k=1$ (needs $Y=4$), $k=2$ ($Y=3$), $k=3$ ($Y=2$), $k=4$ ($Y=1$) — four valid splits, $k=0$ and $k\geq5$ give impossible values of $Y$. Sum: $P(X+Y=5) = \sum_k P(X=k)P(Y=5-k) = 4 \times \left(\tfrac16 \times \tfrac16\right) = \tfrac{4}{36} = \tfrac{1}{9}$. Compare directly by listing outcomes: (1,4),(2,3),(3,2),(4,1) — four out of 36 equally likely pairs, confirming $4/36$.

Worked example 2: two days of normal returns

Day 1 return $X \sim N(0.1\%, 1\%^2)$, day 2 return $Y \sim N(0.1\%, 1.5\%^2)$, independent. Find the distribution of the two-day return $Z = X+Y$. For normals, convolution has a clean shortcut: means add, and variances add (never standard deviations directly). Mean: $0.1\% + 0.1\% = 0.2\%$. Variance: $(1\%)^2 + (1.5\%)^2 = 1 + 2.25 = 3.25$ (in %² units), so the two-day standard deviation is $\sqrt{3.25} \approx 1.803\%$. So $Z \sim N(0.2\%, 1.803\%^2)$. Check the "obvious but wrong" shortcut: adding the standard deviations directly would give $1\% + 1.5\% = 2.5\%$ — noticeably larger than the correct $1.803\%$, because convolution combines spread through variances, not through standard deviations, and the difference between the two grows the more terms you combine.

What this means in practice

Multi-day return distributions, portfolio P&L from summing independent position P&Ls, and aggregate claims in insurance modelling are all convolutions of the individual pieces. The reason the central limit theorem produces a bell curve regardless of the starting distribution is that repeated convolution of any well-behaved distribution with itself smooths toward a normal shape — convolution is the mechanical engine underneath that result. Moment generating functions turn convolution into simple multiplication, which is why they're the standard shortcut for combining many independent variables at once.