normal distribution prove OR disprove E[x]=μ

**nikie1o2** · December 9th 2010, 05:13 PM

normal distribution prove OR disprove E[x]=μ

i believe the claim is true but i'm not sure how to go along to prove it...

**harish21** · December 9th 2010, 06:28 PM

Originally Posted by nikie1o2

normal distribution prove OR disprove E[x]=μ

i believe the claim is true but i'm not sure how to go along to prove it...

Yes it is true.

The pdf of a normal distribution is:

$f(x)=\{\sigma \sqrt{2\pi}\}^{-1} \text{exp}\bigg\{\dfrac{-(x-\mu)^2}{2\sigma^2}\bigg\}$

to calculate the mean, you can evaluate $E[\frac{(X-\mu)}{\sigma}]$ by making a one to one substitution $w=[\frac{(X-\mu)}{\sigma}]$ ..

so,

$E[\frac{(X-\mu)}{\sigma}] = \displaystyle \{\sigma \sqrt{2\pi}\}^{-1} \int_{-\infty}^{\infty} \bigg[\dfrac{(X-\mu)}{\sigma}\bigg] \text{exp}\bigg\{\dfrac{-(x-\mu)^2}{2\sigma^2}\bigg\}\;dx$

$E[\frac{(X-\mu)}{\sigma}] =\displaystyle \underbrace{\{\sqrt{2\pi}\}^{-1} \int_{-\infty}^{\infty}\;w\;\mbox{exp}\bigg\{\dfrac{-w^2}{2}\bigg\}\;dw}_{\text{This is 0. The mean of a standard normal.}}$

The Right hand side of the above equation is zero..since it is the expected value of a standard normal dist. And the proof, like in most Mathematical Statistics books, should be in your book as an example.

$\therefore E[\frac{(X-\mu)}{\sigma}] = 0$

$\bigg[\dfrac{E(X)}{\sigma}\bigg] \;-\; \bigg(\dfrac{\mu}{\sigma}\bigg)=0$

$\therefore E[X]=\mu$

**Guy** · December 10th 2010, 05:02 AM

Originally Posted by harish21

$E[\frac{(X-\mu)}{\sigma}] =\displaystyle \underbrace{\{\sqrt{2\pi}\}^{-1} \int_{-\infty}^{\infty}\;w\;\mbox{exp}\bigg\{\dfrac{-w^2}{2}\bigg\}\;dw}_{\text{This is 0. The mean of a standard normal.}}$

For the sake of completeness, this needs justification. If we don't know the mean of the normal is mu, why do we know that the mean of a standard normal is 0 (or that it even exists?)

**nikie1o2** · December 10th 2010, 06:57 AM

i agree with that also...

**harish21** · December 10th 2010, 07:19 AM

Originally Posted by Guy

For the sake of completeness, this needs justification. If we don't know the mean of the normal is mu, why do we know that the mean of a standard normal is 0 (or that it even exists?)

We are actually trying to show that the mean is mu.

$\displaystyle\{\sqrt{2\pi}\}^{-1} \int_{-\infty}^{\infty}\;w\;\mbox{exp}\bigg\{\dfrac{-w^2}{2}\bigg\}\;dw$

it can be proved that this is equal to zero by using the fact that the above is an odd function. the proof of this should be in the OPs book. If not, they should show their attempt on the problem.

and after finding the mean of $Z$ , a standard random variable, you can use that fact (as done in post #2) to find the mean of another normal random variable X with parameters $\mu \;and\; \sigma$

**Guy** · December 10th 2010, 09:13 AM

Originally Posted by harish21

We are actually trying to show that the mean is mu.

$\displaystyle\{\sqrt{2\pi}\}^{-1} \int_{-\infty}^{\infty}\;w\;\mbox{exp}\bigg\{\dfrac{-w^2}{2}\bigg\}\;dw$

it can be proved that this is equal to zero by using the fact that the above is an odd function. the proof of this should be in the OPs book. If not, they should show their attempt on the problem.

and after finding the mean of $Z$ , a standard random variable, you can use that fact (as done in post #2) to find the mean of another normal random variable X with parameters $\mu \;and\; \sigma$

The fact that it's an odd function doesn't show that the integral exists, because the set being integrated over is not compact...the same argument would imply that the mean of a Cauchy random variable is 0, when in fact the mean doesn't exist.

The easiest way to prove it is to just take the anti derivative and evaluate.

$\displaystyle<br /> \int_{\mathbb{R}} w \exp [-w^2 / 2] \ dw= \left. -\exp[-w^2 / 2] \right| _{-\infty} ^ \infty<br />$

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