Exponential distribution: generalizing mean of the transformation

Question

$X$ is exponentially distributed with parameter $\lambda>0$ and the probability density function is $f_X(x)=\lambda \cdot \exp(\lambda x), \ x>0$

Now consider $X$ as a random variable and the transformed random variable $Y=\exp(x)$

a) find the cdf and pdf of $Y$

$$\text{CDF}= 1-y^{-\lambda}$$ $$\text{PDF}= \lambda \cdot y^{-\lambda-1}$$

This was no problem, I checked with the answer key as well.

b) find the mean of Y

$$\mathbb{E}(Y)=\int_1^{\infty} \lambda \cdot y^{-\lambda}dy=\frac{\lambda}{\lambda-1}$$

Still no problem and correct.

But now I'm curious; does this imply more generally: $$\mathbb{E}(Y^k)=\int_1^{\infty} \lambda \cdot y^{k-\lambda-1}dy=\frac{\lambda}{\lambda-k}?$$

Which again implies that there doesn't exist a moment generating function for $k \ge \lambda, \ \lambda>0$ only?

Thanks in advance!

Answer 1

Yes, ${\mathbb E}(Y^k) = \int_1^\infty \lambda y^{k-\lambda - 1}\ dy$, and this is finite if and only if $k < \lambda$. In particular, $Y$ only has finitely many moments, and there is no moment generating function.

More generally, any random variable whose pdf decays like a power at $\infty$ will only have finitely many moments.