John wants to purchase a bond which will pay him $X$ thousand dollars after two years, where $X$ is equally likely to be any of the numbers in the set $\{0, 1, 2, 3, 4, 5\}$. John believes that the continuously compounded rate of interest, $R$, is independent of $X$ and has a uniform distribution on the interval $(0.04, 0.08)$. The present value of this bond after two years is given by $V = Xe^{-2R}$. Compute the mean of the present value of the bond.
I know:
$F(v) = P(V<=v) = P(Xe^{-2R} <= v)$ ---> $R<= -2ln(\frac{v}{x})$
$D(r) = \frac{r-.004}{0.008-.004} = 25(r-0.04)$
$F(v) = 25(-2ln(\frac{v}{x}) - 0.04)$
the mean is just the Expected value of $F(v)$ for $X = \{0,1,2,3,4,5\}$. Is my work correct$?$
No, there's no need to calculate the density of $V$. Don't make unnecessary work for yourself.
You have $X \sim \mathcal U\{0,1,2,3,4,5\}$, and $R\sim\mathcal U(0.04;0.08)$ and that $X$ and $R$ are independent.
You want to calculate $\mathsf E(X \,\mathsf e^{2R})$. That is:
$$\mathsf E(X\, \mathsf e^{2R}) = \sum_{x=0}^5 x\,p_{\!\lower{0.5ex}X}(x) \int_{0.04}^{0.08} \mathsf e^{2r}f_{\!\lower{0.5ex}R}(r)\operatorname d r$$
Evaluate.