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Suppose that a random variable X has an exponential distribution with rate $\lambda$. Write down the moment generating function for the random variable $-X$. Compute $E(X^n)$, and check calculations of the mean and variance of X.

(The reason for working with $-X$ rather than $X$, is that if t $\geq 0$, (and this is in fact sufficient) then all the integrals are finite.

My attempt so far;

Below I put in $-X$ instead of $X$, and I figure the limits don't change but I can't really convince myself of a good reason why I think that, it's just a gut feeling.

$$M_{-X}(t)=\lambda\int\limits_0^\infty e^{-tX} e^{\lambda X}dx.$$

$$M_{-X}(t)=\lambda\int\limits_0^\infty e^{X(\lambda-t)}dx = \dfrac{\lambda e^{X(\lambda-t)}}{\lambda-t}\Big|_0^\infty$$

$$\dfrac{\lambda e^{X(\lambda-t)}}{\lambda-t}\Big|_0^\infty=\dfrac{\lambda e^{\infty(\lambda-t)}}{\lambda-t}-\dfrac{\lambda e^{0(\lambda-t)}}{\lambda-t}.$$

Assuming $\dfrac{\lambda e^{\infty(\lambda-t)}}{\lambda-t}$ is correct for the first term, I don't know how to evaluate this to get a number. I'm assuming after everything has been evaluated I end up with, $$\dfrac{-\lambda }{\lambda-t}?$$

After this I'm just completely confused. I feel like I lack context over what I'm doing so I can't really think about what I'm supposed to be doing.

Appreciate any help on this, thanks.

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    I wonder if this is a typo, and they really just meant $X$ (not $-X$). Can you write down the density function for $-X$? That is, if $Y=-X$, can you write $f_Y(y)$ in terms of $f_X$? – 2017-02-20
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    It actually has a footnote explaining why it's $-X$, which I'll edit in now. – 2017-02-20
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    $f_Y(y) = \lambda e^{-\lambda y} $ if $ y \geq 0 = \lambda e^{-\lambda -x}$ if $ -x \geq 0 = \lambda e^{\lambda x}$ if $-x \geq 0$? Is that what you mean? – 2017-02-20

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