# E(X) and Var(X) for Normal Dist.

## Homework Statement

Let X be normally distributed with the paremeters 0 and σ2. Find:
a. E(X2)
b. E(aX2+b)

## Homework Equations

E(X) = $\int_{-\infty}^{\infty} \! xf(x) \mathrm{d} x$

E(X2) = $\int_{-\infty}^{\infty} \! x^2f(x) \mathrm{d} x$

E(aX+b) = $aE(X)+b$

The normal distribution with paremeters 0 and σ2 = $( \frac{1}{\sigma \sqrt{2 \pi }}) e^{ \frac{-x^2}{2 \sigma ^2}}$

## The Attempt at a Solution

a. $E(X^2)= \frac{1}{ \sigma \sqrt{2 \pi }} \int_{-\infty}^{\infty} \! x^2e^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x$

I don't knowhow to solve this integral. I think there might some tricks involved using the fact that this is a distribution function.

b. I just need to find E(X) which I start by setting up: $E(X)= \frac{1}{ \sigma \sqrt{2 \pi }} \int_{-\infty}^{\infty} \! xe^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x$

I don't know how to solve this integral either.

I also don't know how to solve the simpler integral $\int e^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x$

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LCKurtz
Homework Helper
Gold Member
Try integration by parts with
$$u = x,\, dv = xe^{-\frac {x^2}{2\sigma^2}}$$
Then you should be able to use the fact that the first moment is 0 and the fact that the normal distribution function is a distribution function.

I like Serena
Homework Helper
Hi Arcana!

The integral $\int_{-\infty}^x e^{-t^2}dt$ is proven to be unsolvable (with standard functions).

That's why they introduced a new function to deal with this (the cdf of the standard normal distribution):
$$\Phi(x) = {1 \over \sqrt{2\pi}} \int_{-\infty}^x e^{-t^2 \over 2}dt$$

$\Phi$ has the property that $\Phi(+\infty) = 1$, but to find any regular function value it needs to be approximated numerically.

The trick for your problem is to use partial integration (as LCKurtz suggested).
Either you can solve directly, or you can reduce your formula to something like $\int_{-\infty}^{+\infty} e^{-t^2 \over 2}dt$ and use that it is $\sqrt{2\pi}$.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Let X be normally distributed with the paremeters 0 and σ2. Find:
a. E(X2)
b. E(aX2+b)

## Homework Equations

E(X) = $\int_{-\infty}^{\infty} \! xf(x) \mathrm{d} x$

E(X2) = $\int_{-\infty}^{\infty} \! x^2f(x) \mathrm{d} x$

E(aX+b) = $aE(X)+b$

The normal distribution with paremeters 0 and σ2 = $( \frac{1}{\sigma \sqrt{2 \pi }}) e^{ \frac{-x^2}{2 \sigma ^2}}$

## The Attempt at a Solution

a. $E(X^2)= \frac{1}{ \sigma \sqrt{2 \pi }} \int_{-\infty}^{\infty} \! x^2e^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x$

I don't knowhow to solve this integral. I think there might some tricks involved using the fact that this is a distribution function.

b. I just need to find E(X) which I start by setting up: $E(X)= \frac{1}{ \sigma \sqrt{2 \pi }} \int_{-\infty}^{\infty} \! xe^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x$

I don't know how to solve this integral either.

I also don't know how to solve the simpler integral $\int e^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x$
Don't you mean $E(a X^2 + b) = a E(X^2) + b$? I am surprised that you have not seen the standard result that $\mbox{Var}(X) = E(X^2) - (EX)^2$, which is true for _any_ random variable having finite mean and variance. It ought to be in your textbook or course notes.

RGV

I don't remember why I thought finding E(X) was going to help. As for E(X2), here is my attempt. And I'm sure I'm breaking tons of rules here.

$$E(X^2) = \frac{1}{ \sigma \sqrt{2 \pi }} \int_{-\infty}^{\infty} \! x^2e^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x$$
$$u=x^2$$ $$\mathrm{d} u=2x \mathrm{d} v$$ $$v= \sqrt{2\pi }$$ $$dv= e^{-\frac {x^2}{2\sigma^2}} \mathrm{d} x$$

$$\frac{1}{ \sigma \sqrt{2 \pi }} \int_{-\infty}^{\infty} \! x^2e^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x =$$
$$\frac{1}{ \sigma \sqrt{2 \pi }}( x^2 \sqrt{2\pi } ^{-\infty }_{\infty } - \sqrt{2\pi } \int_{-\infty}^{\infty} \! 2x \mathrm{d} x )$$

Am I messing it up? I don't think I'm on the right track.

Last edited:
For $E[X^2]$, do integration by parts by putting

$u=x,~v=xe^{\frac{-x^2}{2\sigma^2}}$

I like Serena
Homework Helper
Let's see.
We want to apply partial integration, which is:
$$\int u dv = u v - \int v du$$

I hope you don't mind that I point out a few rules that you are breaking. :shy:
Perhaps we can fix that.

I don't remember why I thought finding E(X) was going to help. As for E(X2), here is my attempt. And I'm sure I'm breaking tons of rules here.

$$E(X^2) = \frac{1}{ \sigma \sqrt{2 \pi }} \int_{-\infty}^{\infty} \! x^2e^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x$$

$$u=x^2$$$$\mathrm{d} u=2x \mathrm{d} v$$
With your choice for u this should be: $\mathrm{d} u=2x \mathrm{d} x$

$$v= \sqrt{2\pi }$$
With this choice for v, we would get: $dv = 0$.
That can't be right.

$$dv= e^{-\frac {x^2}{2\sigma^2}} \mathrm{d} x$$
With this choice for dv, I don't see how you can figure out v, since this expression does not have an anti-derivative (expressed in standard functions).

$$\frac{1}{ \sigma \sqrt{2 \pi }} \int_{-\infty}^{\infty} \! x^2e^{ \frac{-x^2}{2 \sigma ^2}} \mathrm{d} x =$$
$$\frac{1}{ \sigma \sqrt{2 \pi }}( x^2 \sqrt{2\pi } ^{-\infty }_{\infty } - \sqrt{2\pi } \int_{-\infty}^{\infty} \! 2x \mathrm{d} x )$$
There should be a vertical line in there, and the limits should be switched around.
But let's take care of that when the rest is in order.

Am I messing it up? I don't think I'm on the right track.
Only a little bit.
It needs to be tweaked a bit to get on the right track.

I recommend using:
$$u = x$$$$dv = x e^{-\frac {x^2}{2\sigma^2}} \mathrm{d} x$$

Of course this means that you have to integrate dv to find v...

Hooray! The problem is solved! Thanks team :)

LCKurtz