Cnditional expectation of exponential on UDF

Question

Let $X$ be an exponential random variable with $\lambda = 5$ and $Y$ a uniformly distributed random variable on $(-3,X)$. Find $\mathbb{E}[Y]$.

I tried to solve it by just integrating $f(y)$ from $(-3, x)$, but that doesn't seem to be working.

Answer 1

The (conditional) expectation of $Y$ is $\frac{X+3}{2}$. So you want $$\int_0^\infty \frac{x+3}{2}\cdot 5e^{-5x}\,dx.$$ Here I am assuming that by the exponential with parameter $\lambda$ you mean that the density function is $\lambda e^{-\lambda x}$, $x\gt 0$. More infrequently, some people think of it as having density $\frac{1}{\lambda}e^{-x/\lambda}$.

Answer 2

You know that $\mathsf E(X)=\lambda^{-1} = 1/5$ and $\mathsf E(Y\mid X)=(X+3)/2$ , from what was given.

So you may make good use of the tower rule (aka the Law of Iterated Expectation). $$\mathsf E(Y)=\mathsf E(\,\mathsf E(Y\mid X)\,)$$