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Thread: Central Limit Theorem

  1. #1
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    Central Limit Theorem

    Let X1,X2, be a sequence of independent N(0, 8) distributed random variables. For n=1,2, let Yn be the random variable defined by



    Yn=X^2(subscript)1+⋯+X^2(subscript)n.
    Use the central limit theorem rule of thumb to approximate P(Y25>216). You can use that E[X^4(subscript)i]=192.

    I'm really struggling with how to do this question, I tried to use the formula for Var(x)=e[x^2] - [E[x]]^2
    and tried standardising using Zn = (Xn(bar) - E[x])/ sigma/root(n) but it just doesn't seem to be working. Any help and explanations would be greatly appreciated.
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  2. #2
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    Re: Central Limit Theorem

    Quote Originally Posted by cs1632 View Post
    Let X1,X2, be a sequence of independent N(0, 8) distributed random variables. For n=1,2, let Yn be the random variable defined by



    Yn=X^2(subscript)1+⋯+X^2(subscript)n.
    Use the central limit theorem rule of thumb to approximate P(Y25>216). You can use that E[X^4(subscript)i]=192.

    I'm really struggling with how to do this question, I tried to use the formula for Var(x)=e[x^2] - [E[x]]^2
    and tried standardising using Zn = (Xn(bar) - E[x])/ sigma/root(n) but it just doesn't seem to be working. Any help and explanations would be greatly appreciated.
    ok, you've got a sum of the squares of iid normal rvs each $\sim N(0,\sigma^2)$.

    The first step is to determine the mean and variance of each of these.

    We find that

    $E[Y_1] = \sigma^2$

    $Var[Y_1] = 2\sigma^4$

    so by the central limit theorem

    $\sqrt{25}\left(\left(\dfrac{1}{25} \displaystyle{\sum_{k=1}^{25}}~X^2_k\right)-\sigma^2\right) \Rightarrow N(\sigma^2, 2\sigma^4)$

    you should be able to complete the algebraic massage needed to answer your problem.
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  3. #3
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    Re: Central Limit Theorem

    Quote Originally Posted by romsek View Post
    ok, you've got a sum of the squares of iid normal rvs each $\sim N(0,\sigma^2)$.

    The first step is to determine the mean and variance of each of these.

    We find that

    $E[Y_1] = \sigma^2$

    $Var[Y_1] = 2\sigma^4$

    so by the central limit theorem

    $\sqrt{25}\left(\left(\dfrac{1}{25} \displaystyle{\sum_{k=1}^{25}}~X^2_k\right)-\sigma^2\right) \Rightarrow N(\sigma^2, 2\sigma^4)$

    you should be able to complete the algebraic massage needed to answer your problem.
    my mistake

    $\sqrt{25}\left(\left(\dfrac{1}{25} \displaystyle{\sum_{k=1}^{25}}~X^2_k\right)-\sigma^2\right) \Rightarrow N(0, 2\sigma^4)$
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