Expected value of function of exponential random variable

Question

Let $X \sim Exp(\lambda)$ (exponential distribution) and $Y=X^a$. For what values of $a\in R$ is $E[Y]< \infty$?

I computed $E[Y] = \lambda \int x^a e^{-\lambda x}dx$. How to proceed? Does the answer depend on $\lambda$?

Answer 1

Well, using the definition of Gamma function, we can see that \begin{equation} \mathbb{E}[Y] =\frac{\Gamma(a+1)}{\lambda^{a}}. \end{equation}

Next, using Prym's decomposition of Gamma function, we know that \begin{equation} \Gamma(a)= \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(z+n)}+\int_{1}^{\infty} x^{a-1}{\rm e}^{-x}{\rm d} x. \end{equation} Hence, $\Gamma(a)$ has simple poles on negative integer.

Moreover, for negative non-integer numbers, you can easily see that the series is absolutely convergent (take a look at the definition of exponential series).

Answer 2

HINT

You are correct in that $$\mathbb{E}[Y] = \lambda \int_0^\infty x^a e^{-\lambda x}dx$$ and since you need $\mathbb{E}[Y] < \infty$, you are looking for a range of $a$ for which the integral converges.

Let's first play with a simple case of $\lambda = 1$, we are looking for $\int_0^\infty x^a e^{-x} dx$, which has 2 potential problems, at $x = 0$ if $a < 0$ and at $x \to \infty$. So we can write $$ \int_0^\infty x^a e^{-x} dx = \int_0^1 x^a e^{-x} dx + \int_1^\infty x^a e^{-x} dx. $$

Let's deal with the problem at $x \to \infty$ first, so look at the second integral. For any value of $a$ and $x$ large enough, you can bound the integrand by $1/x^2$ (can you prove it?) and the integral would converge.

For the first integral, we have a problem when $a < 0$. Can you finish the analysis for this case and generalize for other $\lambda > 0$?

UPDATE

For the second integral, if we want to bound $$ x^ae^{-x} < x^{-2} \iff \frac{x^{2+a}}{e^x} < 1 $$ What is the limit of the LHS as $x \to \infty$?

Answer 3

Hint:

(1) $a\leq-1$

$\int_0^{1} x^{a}e^{-x} dx$ diverges.

Since For $0<x\leq 1$ $$e^{-x}\geq e^{-1} \hspace{.5cm} and \hspace{.5cm} x^{a} \geq x^{-1}$$ so $$\int_0^{1} x^{a}e^{-x} dx \geq \int_0^{1} e^{-1} x^a dx \geq \int_0^{1} e^{-1} \frac{1}{x} dx $$ and $\int_0^{1} \frac{1}{x} dx $ diverges.

(2) For $a>-1$

$$\int_0^{+\infty} x^{a}e^{-x} dx=\int_0^{+\infty} x^{(a+1) -1}e^{-x} dx=\Gamma(a+1)$$ see Incomplete_gamma_function.

Gamma distribution is also useful. For $k>0$ (equal with a>-1) and $\lambda>0$

$$\int_0^{+\infty} \frac{\lambda^k }{\Gamma(k)} x^{k-1}e^{-\lambda x} dx=1$$

so for any $k>0$

$$\int_0^{+\infty} x^{k-1}e^{-\lambda x} dx= \frac{\Gamma(k)}{\lambda^k}$$