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Math Help - Normal Distributions and Integrals

  1. #1
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    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?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by clipperdude21 View Post
    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:

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

    Put:

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


    Now consider:

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

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

    Hence: I(u)=u

    RonL
    Last edited by CaptainBlack; November 10th 2007 at 12:04 PM.
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  3. #3
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    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!
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