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Thread: Median and Mean of Independent Random Variables

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    Median and Mean of Independent Random Variables



    Hello,

    Here is the question:

    Let X_1,X_2, . . . ,X_n be random variables with the density \frac{1}{1+x^2} for x \in ({-\infty},{\infty}) ; suppose further that these random variables are independent.

    a. What is the median of each of the X_j?

    b. Do you expect that the mean of these random variables will be approximately normal for large n ? Why or why not?

    A walk through to the solution would be greatly appreciated!

    Thank you!

    Matt.
    Last edited by Dason; 12-22-2010 at 11:58 AM.

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    RotParaTon
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    Re: Median and Mean of Independent Random Variables

    What you have posted isn't actually a density. If you multiply by \frac{1}{\pi} it is the standard cauchy distribution though. The wikipedia page answers all your questions. I don't know how much of a walkthrough it gives though.

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    Re: Median and Mean of Independent Random Variables

    For question a) it is my understanding that the integral of the density needs to be set to 0.5, and the answer is 0. Giving arctan(x) = 0.5 -> x = tan(0.5) but that does not equal 0?

    Any ideas?

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    Re: Median and Mean of Independent Random Variables

    Is that really the CDF though? Recall the 1/pi I mention earlier. Also recall that for something to be a CDF the limit as you go to infinity should be 1 and the limit as you go to negative infinity should be 0. You may need to add or subtract an integration constant to make that a CDF.

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    Re: Median and Mean of Independent Random Variables

    The median is zero if the density you mentioned is multiplied by 1/pi, as Dason pointed out.
    For the sample mean, according to Central limit theorems (note the "s" here there is no single CLT. There are different "versions"), it does not follow normal dist since CLTs require the underlying distributions of those rv have first and second moments. Cauchy dist does not even hv a finite first moment.

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    Re: Median and Mean of Independent Random Variables


    Note that you can't really use the CLT to show that something doesn't converge to a normal since as far as I know they all give sufficient but not necessary conditions for the convergence to take place. Somebody might try to justify the use here but it's actually not too bad to just show directly that the sample mean follows a cauchy distribution as well.
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