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Thread: integration to find expectation

  1. #1
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    integration to find expectation

    Hi

    I need to work out this integral $\displaystyle $$\int^{\infty}_{0}a^2y^2e^{\frac{-a^2y^2}{2}}dy$$$. I have already multiplied the expression by y, since i'm trying to work out E(Y).

    Now I tried to look at the Gamma and Beta functions, but I can't recognise how they would work here, mainly because of the exponential part contains a $\displaystyle y^2$.

    Any help is much appreciated
    Last edited by cooltowns; Oct 26th 2017 at 08:16 AM.
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  2. #2
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    Re: integration to find expectation

    Rewrite this as 1/2 the expected value of $y^2$ when $y$ is zero mean normally distributed with standard deviation $\dfrac 1 a$

    You'll probably also have to multiply by a constant to the get the proper form.
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  3. #3
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    Re: integration to find expectation

    Hi Romsek

    my PDF was actually this $\displaystyle $$\int^{\infty}_{0}a^2ye^{\frac{-a^2y^2}{2}}dy$$$

    hence i wrote $\displaystyle $$\int^{\infty}_{0}a^2y^2e^{\frac{-a^2y^2}{2}}dy$$$ because I wanted E(Y).
    Last edited by cooltowns; Oct 26th 2017 at 08:23 AM.
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  4. #4
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    Re: integration to find expectation

    Quote Originally Posted by cooltowns View Post
    Hi Romsek

    my PDF was actually this $\displaystyle $$\int^{\infty}_{0}a^2ye^{\frac{-a^2y^2}{2}}dy$$$

    hence i wrote $\displaystyle $$\int^{\infty}_{0}a^2y^2e^{\frac{-a^2y^2}{2}}dy$$$ because I wanted E(Y).
    Whatever. The method I suggested will allow you to compute the integral.
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