Expectation of a function of a uniform random variable

Question

Let $X$ be a uniform random variable with the pdf of $f_{X}(x) = \frac{1}{\alpha}, 0 \le x \le \alpha$ where $\alpha$ is a real number. The goal is to find a closed-form expression of $\mathbb{E}_{X} \left[ e^{-\lambda X} \frac{(\lambda X)^{n_0}}{k!} \right]$, where $\lambda > 0$, $n_0 > 0$, and $k > 0$ are all constants.

Some thoughts: I see the insider of the expected value form a Gamma pdf, which might be used to derive $\mathbb{E}_{X} \left[ e^{-\lambda X} \frac{(\lambda X)^{n_0}}{k!} \right]$.

Or, if it is possible to assume that $e^{-\lambda X}$ and $\frac{(\lambda X)^{n_0}}{k!}$ are independent, I would have proceeded to splitting the expectation into $\mathbb{E}_{X} \left[ e^{-\lambda X} \frac{(\lambda X)^{n_0}}{k!} \right] = \mathbb{E}_{X} \left[ e^{-\lambda X} \right] \mathbb{E} \left[ \frac{(\lambda X)^{n_0}}{k!} \right]$.

However, I am unsure either of these is violating any mathematical generality.

Answer 1

$$\mathbb{E}_X\left[e^{-\lambda X} \frac{(\lambda X)^{n_0}}{k!}\right] = \frac{1}{\alpha} \int_0^\alpha e^{-\lambda x}\frac{(\lambda x)^n_0}{k!} \ \text{d}x$$ with the substitution $u = \lambda x$ gives $$ \mathbb{E}_X\left[e^{-\lambda X} \frac{(\lambda X)^{n_0}}{k!}\right] = \frac{1}{\alpha \lambda k!}\int_0^{\frac{\alpha}{\lambda}}e^{-u}u^{n_0} \ \text{d}u $$ Can you take it from here?

Answer 2

Since both terms depend on $X$, they are not independent. However for the given distribution, the expression you want is straightforward, even if a bit messy.

$A=E_X(e^{-\lambda X}\frac{(\lambda X)^n}{k!})=\frac{1}{\alpha k!}\int_0^\alpha e^{-\lambda x} (\lambda x)^ndx$

$A=\frac{1}{ak!}[\frac{n!}{\lambda}-\frac{e^{-a\lambda}}{\lambda}\sum_{j=0}^n\frac{n!}{(n-j)!}(a\lambda)^{n-j}]$