# Thread: Normal Distributions and Integrals

1. ## Normal Distributions and Integrals

The question says that suppose f(x) is a density function of a normal distribution with mean u and standard dev o(sigma). Sow that u= integral from -infinity to infinity of x(fx)dx.

= f(x)

What I did was substitue 0 for u and 1 for sigma since its a normal distribution. THen took the integral using substitution and got the integral =0. I then said since u=o and the integral =0 then u=integral. Is that correct?

2. Originally Posted by clipperdude21
The question says that suppose f(x) is a density function of a normal distribution with mean u and standard dev o(sigma). Sow that u= integral from -infinity to infinity of x(fx)dx.

= f(x)

What I did was substitue 0 for u and 1 for sigma since its a normal distribution. THen took the integral using substitution and got the integral =0. I then said since u=o and the integral =0 then u=integral. Is that correct?
You want to show that:

$\displaystyle u = \int_{-\infty}^{\infty} x \frac{1}{\sigma \sqrt{2 \pi}} e^{-(x-u)^2/(2\sigma^2)}~dx$

Put:

$\displaystyle I(u) = \int_{-\infty}^{\infty} x \frac{1}{\sigma \sqrt{2 \pi}} e^{-(x-u)^2/(2\sigma^2)}~dx$

Now consider:

$\displaystyle I(u)-u=\int_{-\infty}^{\infty} (x-u) \frac{1}{\sigma \sqrt{2 \pi}} e^{-(x-u)^2/(2\sigma^2)}~dx$$\displaystyle =\frac{1}{\sigma \sqrt{2 \pi}} \left[ -2 \sigma^2 e^{-(x-u)^2/(2\sigma^2)} \right]_{x=-\infty}^{\infty}=0$

(the first step is using the fact that $\displaystyle u \int_{-\infty}^{\infty} f(x)~dx = u$)

Hence: $\displaystyle I(u)=u$

RonL

3. Hi Thanks Captain Black... I have one question though. When you substracted the u from each side. How did you know you could put it inside the integral and make it integral of (x-u)f(x)dx?

Thanks!