# Expected value of a function of a jointly distributed random variable?

Imagine you have a jointly distributed random variable $X(t)$, where $X(t)=Ae^{-Bt}$, where $A$~$\text{Exponential} (1/2)$ and $B$~$\text{Uniform} (0,1)$ such that $A$ and $B$ are independent.

How would you find the expected value of $X(t)$?

I've been trying to use properties of expectations to solve it - ie. saying: $E(Y(t))=E(Ae^{-Bt})=E(A)E(e^{-Bt})$. And we're allowed to assume we know $E(A)=2$ and $E(B)=1$.

So $E(1)E(e^{-t})$

But I'm not sure this is the correct way to do it. And I don't know where to go after this.

First of all if $A\sim \exp(1/2)$ then $E[A]=(1/2)^{-1}=2$ (+) (from you question it seems like you are a bit unsure on this) applying this and using your work we only need to find $E[e^{-Bt}]$. For this we can use this which helps find the expectation of any (nice) function of a random variable. In our case let $t$ be given and $g(x)=e^{-x t}$ then by the link $E[g(B)]=\int_{-\infty}^\infty g(x) f_B(x) dx$ where $f_B$ is the density of $B$. This will be $\int_0^1 e^{-t\cdot x}dx$ (why?) which you can easily calculate and insert into your expression together with (+) to yield the result. As a "funny" side note $E[e^{-X\cdot t}]$ is called the characteristic function and is really nice :)