这是一份ed.ac爱丁堡格大学MATH11154作业代写的成功案

$$
\mathbb{P}(X+Y \leq a)=\int_{-\infty}^{\infty} F_{X}(a-y) f_{Y}(y) \mathrm{d} y
$$
Taking the derivatives of both sides with respect to $a$, we have
$$
f_{X+Y}(a)=\int_{-\infty}^{\infty} f_{X}(a-y) f_{Y}(y) \mathrm{d} y=f_{X} * f_{Y}(a)
$$
We used the fact that $F_{X}$ is sufficiently smooth.
A more mathematically rigorous proof goes as follows: Since the joint probability density function of $(X, Y)$ is equal to $f_{X}(x) f_{Y}(y)$, we have
$$
\mathbb{P}(X+Y \leq a)=\int_{-\infty}^{\infty} \int_{-\infty}^{a-y} f_{X}(x) f_{Y}(y) \mathrm{d} x \mathrm{dy}
$$
Differentiating both sides with respect to $a$, we obtain
$$
f_{X+Y}(a)=\int_{-\infty}^{\infty} f_{X}(a-y) f_{Y}(y) \mathrm{dy}
$$

MATH11154 COURSE NOTES ：

First, consider the case when $h$ is monotonically increasing. Since
$$
\operatorname{Pr}(a \leq X \leq x)=\operatorname{Pr}(h(a) \leq Y \leq h(x)),
$$
we have
$$
\int_{a}^{x} f_{X}(x) \mathrm{d} x=\int_{h(a)}^{h(x)} f_{Y}(y) \mathrm{d} y
$$
By differentiating with respect to $x$, we have
$$
f_{X}(x)=f_{Y}(h(x)) h^{\prime}(x)
$$
Now substitute $x=h^{-1}(y)$. For the case when $h$ is decreasing, note that
$$
\int_{a}^{x} f_{X}(x) \mathrm{d} x=\int_{h(x)}^{h(a)} f_{Y}(y) \mathrm{d} y=-\int_{h(a)}^{h(x)} f_{Y}(y) \mathrm{d} y
$$