Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ and $Y$ be normal random variables and $Z$ be s.t. $Z=aX+bY$. Then, find $M_Y(t)=E(e^{tY})$ and $M_Z(t)=E(e^{tZ})$.

Attempt: So I know that for normal random variable $Y$, $E(Y)=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-(x-\mu)^2/2\sigma^2}dx$. So for the marginal distribution of $Y$, it should be something like $E(e^{tY})=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}e^{tY}e^{-(e^{tY}-\mu)^2/2\sigma^2}dx$... kind of stuck at thi s point.

Edit: I just had an idea -- shouldn't the marginal distribution of $Y$ just be the probability distribution of $Y$? -- $E(Y)=\frac{1}{\sqrt{2\pi}\sigma_{Y}}xe^{-(x-\mu)^2/2\sigma_Y^2}$?

share|cite|improve this question
up vote 1 down vote accepted

Let $X$ be normal, mean $\mu$, variance $\sigma^2$, and let $Y$ be normal, mean $\nu$, variance $\tau^2$.

Since $X$ and $Y$ are independent, $aX+bY$ is normal, mean $a\mu+b\nu$, and variance $a^2\mu+b^2\nu$. This fact is quite useful, and will help you ompute the moment-generating function $M_Z(t)$ of $Z$.

Now we find a formula for the moment generating function of $X$. We want to find the expectation of $e^{tX}$ (we are computing this one, instead of doing the essentially identical calculation of $E(e^{tY})$, because of a preference for the letter $X$.) We have $$E(e^{tX})= \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{tx}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx.$$ It remains to evaluate this integral. We do the calculation in two steps, though it could be done in one. Make the substitution $w=\dfrac{x-\mu}{\sigma}$. Routine use of the substitution process shows that our integral is $$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{t(\mu+\sigma w)}e^{-\frac{w^2}{2}}dw.$$ We can simplify this to $$e^{t\mu}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{w^2}{2} +t\sigma w}dw.$$ Look at the exponent in the above integral. It is $-\frac{w^2}{2}+t\sigma w$, which is $-\frac{1}{2}(w^2-2t\sigma w)$. Completing the square, we find that the exponent is $$-\frac{1}{2}(w-t\sigma)^2+\frac{t^2\sigma^2}{2}.$$ We can bring the
$e^{\frac{t^2\sigma^2}{2}}$ part "out" of the integral, and find that our integral is $$e^{t\mu+\frac{t^2\sigma^2}{2}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{(w-t\sigma)^2}{2}}dw.$$ It's almost over! Make the substitution $z=w-t\sigma$. We get that our integral is $$e^{t\mu+\frac{t^2\sigma^2}{2}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz.$$ The remaining integral has value $1$, since it is the area under the (standard) normal curve. We conclude that $$E(e^{tX})=e^{t\mu+\frac{t^2\sigma^2}{2}}.$$

You will not have to sweat like this to find $E(e^{tZ})$. You can probably just borrow the result we have just proved, and replace $\mu$ by $a\mu+b\nu$, and replace $\sigma^2$ by $a^2\sigma^2+b^2\tau^2$.

But maybe not. It all depends on whether it has been proved, or you can take for granted, that a linear combination of independent normals is normal.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.