Convolution sums

Suppose $X$ and $Y$ are Independent r.v.s and we know their distributions. We want to find the distribution of $T=X+Y$.

Discrete

The PMF of the sum $T=X+Y$ is given by

$$\begin{array}{rl}P(T=t)& =\sum _{x\in \text{Support}(X)}P(X=x)\cdot P(Y=t-x)\\ & =\sum _{y\in \text{Support}(Y)}P(Y=y)\cdot P(X=t-y)\end{array}$$

To prove, consider $P(T=t)=P(X+Y=t)$. By LOTP and WLOG, this is equivalent to

$$P(X+Y=t)=\sum _{x\in \text{Support}(X)}P(X+Y=t\mid X=x)\cdot P(X=x)$$

Consider $P(X+Y=t\mid X=x)$. We’re given that $X=x$, therefore this is equivalent to saying $P(x+Y=t\mid X=x)=P(Y=t-x\mid X=x)$. Finally, we know $X$ and $Y$ are independent, so it simplifies down to $P(Y=t-x)$.

Continuous

The PDF of $T=X+Y$ is given by

$$\begin{array}{rl}{f}_T(t)& ={\int }_{-\infty }^{\infty }{f}_X(x)\cdot f_Y(t-x)dx\\ & ={\int }_{-\infty }^{\infty }{f}_Y(y)\cdot f_X(t-y)dy\end{array}$$