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

$P(T=t) =x∈Support(X)∑ P(X=x)⋅P(Y=t−x)=y∈Support(Y)∑ P(Y=y)⋅P(X=t−y) $To prove, consider $P(T=t)=P(X+Y=t)$. By LOTP and WLOG, this is equivalent to

$P(X+Y=t)=x∈Support(X)∑ P(X+Y=t∣X=x)⋅P(X=x)$Consider $P(X+Y=t∣X=x)$. We’re given that $X=x$, therefore this is equivalent to saying $P(x+Y=t∣X=x)=P(Y=t−x∣X=x)$. 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

$f_{T}(t) =∫_{−∞}f_{X}(x)⋅f_{Y}(t−x)dx=∫_{−∞}f_{Y}(y)⋅f_{X}(t−y)dy $