Show that
E(X)=μ<br><br>thats all the information i've been given.<br>I dont know where to start.<br>
First, note that the pdf of $\displaystyle \mathcal{N}\left(\mu,\sigma^2\right)$ is $\displaystyle f\!\left(x\right)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$
Now, $\displaystyle E\left(x\right)=\int_{-\infty}^{\infty}xf\!\left(x\right)\,dx=\frac{1}{\s qrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx$
Making the substitution $\displaystyle u=\frac{x-\mu}{\sigma}\implies \sigma\,du=\,dx$.
Thus, we have $\displaystyle \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx\xrightarrow{u=\left(x-\mu\right)/\sigma}{}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\sigma u+\mu\right)e^{-\frac{1}{2}u^2}\,du$ $\displaystyle =\frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du+\frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du$
For the first integral:
$\displaystyle \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du$
Let $\displaystyle z=\tfrac{1}{2}u^2\implies \,dz=u\,du$
Thus, $\displaystyle \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2} {}\frac{\sigma}{\sqrt{2\pi}}\int_{\infty}^{\infty} e^{-z}\,dz=\frac{\sigma}{\sqrt{2\pi}}\cdot0=0$.
Now for the second integral:
$\displaystyle \frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du$
Recall that the Gaussian integral $\displaystyle \int_{0}^{\infty}e^{-ax^2}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\implies\ int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}} $.
So it follows that
$\displaystyle \frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du=\frac{\mu}{\sqrt{2\pi}}\cdot\s qrt{\frac{\pi}{1/2}}=\frac{\mu}{\sqrt{2\pi}}\cdot\sqrt{2\pi}=\mu$.
Therefore, $\displaystyle E\left(x\right)=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx=\mu$.
Does this make sense?
I didn't integrate the expected value. That's how the expected value is defined for continuous distributions (like the normal):
$\displaystyle E\left(X\right)=\int_{-\infty}^{\infty}xf\!\left(x\right)\,dx$, where $\displaystyle f\!\left(x\right)$ is the pdf of the distribution.
Here's a site that evaluated the integral the way I did: The Normal Distribution
Note that
Thus, $\displaystyle \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2} {}\frac{\sigma}{\sqrt{2\pi}}\int_{\infty}^{\infty} e^{-z}\,dz=\frac{\sigma}{\sqrt{2\pi}}\cdot0=0$.
is a lot of work.
The integrand is an odd function so the integral must be zero.
I just commented on that in my review of that calc book.
Noticing that a function is even or odd can save some work.
As for
$\displaystyle \frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du$
I first show that the normal is a valid density with that polar coordinate trick (multiplying it by itself)
and then we have
$\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du=1$
since this is a valid density of a N(0,1).
After the substitution $\displaystyle z=\tfrac{1}{2}u^2$, the limits of integration became positive (and the same)!
So it follows from integration properties that $\displaystyle \int_a^a f\!\left(x\right)\,dx=0$.
That's why $\displaystyle \int_{\infty}^{\infty}e^{-z}\,dz=0$