$X$ and $Y$ are ind. exponentially dist. ran. variables w/para. $\beta_1$ and $\beta_2$. Let $U=X+Y$, verify that $f_u(u)= \int_0^u f_{xy}(u-v,v)dv$.

Question

I am a little lost with transformations with exponential distributions, any help would be much appreciated!

The given hint is $0<x<infty$ and $x=u-v$

Answer 1

We have $X \geq 0$ and $Y \geq 0$.

The probability that $0 \leqslant U =X+Y \leq \hat{u}$ is obtained by integrating the joint probability density function $f_{xy}$ over the triangular region in the $xy$-plane with vertices at $(0,0)$,$(0,\hat{u})$, and $(\hat{u},0)$:

$$P(U \leq \hat{u}) = \int_0^{\hat{u}}f_u(u)du=\int_0^{\hat{u}}\int_0^{\hat{u}-x}f_{xy}(x,y)\,dy \, dx.$$

Make the change of variables $x = u-v, y=v$.

The Jacobian is $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=1$ and the region transforms to a triangular region in the $uv$-plane with vertices at $(0,0)$, $(\hat{u},0)$ and $(\hat{u},\hat{u})$.

Then

$$\int_0^{\hat{u}}f_u(u)du=\int_0^{\hat{u}}\int_0^{u}f_{xy}(u-v,v)\,dv \, du,$$

and

$$f_u(u) = \frac{\partial}{\partial \hat{u}}\int_0^{\hat{u}}\int_0^{u}f_{xy}(u-v,v)\,dv \, du \\=\int_0^{u}f_{xy}(u-v,v)\,dv.$$