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A friend thinks if he performs the following steps, he will develop a new distribution called the World's Greatest distribution. Suppose $X_1,X_2, \ldots ,X_n$ are iid $N(\mu, \sigma^2)$. He takes each one of these random variables, subtracts $\mu$ from each of them, and divides each of them by $\sigma^2$. He then squares each one of these new random variables, takes their sum and makes a new random variable he calls $D$. Then he adds $4$ $\mathrm{Gamma}(3,2)$ random variables, $4$ $\mathrm{Weibull}(1,2)$ random variables, and $9$ $\mathrm{Exponential}(2)$ random variables to $D$ where all of these random variables including $D$ are mutually independent. He calls this new random variable $U$. Now he divides $U$ by a constant $C$ and then divides that result by a chi-squared random variable with $100$ degrees of freedom divided by $100$. This chi-squared random variable is independent of $U$. He calls this new random variable $M$. After doing all of this work, he then takes the reciprocal of $M$ and calls this final random variable $B$. Your job is to check his work and see if this resulting random variable is something you know already. You find out that it is. Thus, you go and tell him the news. What distribution is it? (Note: You will need to determine the constant $C$ that divides $U$.)

For this I see that I have $N$ normal distributions and multiply by the moment generating functions of the $4$ gammas, $4$ Weibulls and $9$ exponentials which means I only add up the exponents because mgf is of the same form. Thus I got $\bigl(\frac{1}{1-2t})^{\frac{n}{2}+25}$. Getting a little lost after that.

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sub(s)tracts from each of them WHAT? and divides each of them by WHAT? (Pranks are all right with me, but the art of pranking has some rules and pranks with missing pieces are just silly.) –  Did Dec 15 '12 at 10:35
    
Sorry, I transferred this from a pdf and lost the variables. They are replaced now. Just a simple mistake. Don't see how this would be meant as a prank. –  user45185 Dec 16 '12 at 3:24
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Prank refers to the fact that random variable B is obtained through a series about 10 different steps, uncorrelated, chosen with no apparent logic, some of these steps nontrivial to translate into the distribution obtained. This, joined to the lack of serious mathematical justification for this construction and to the tongue-in-cheek (I hope) invocation of a World's Greatest distribution makes me suspect that some kind of social experiment might be at work here, rather than a honest math question. Is it? –  Did Dec 16 '12 at 8:31

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Hmmm...Interesting Question. I think that you meant, instead of $\sigma^2$ that you wanted $\sigma$. Is that correct? (I assume it is and the following work I used $\sigma$.

I just worked it out, and when you multiply the mgf's of the Weibull, Gamma, exponential and the Chi-Squared with n-degrees of freedom, you do indeed get the mgf that you had. Hint: Make the exponent so that n and 25 share a common denominator. Then look at the continuous distribution mgf's and see if any of them have that form but to a power. You will see that two of the continuous distribution's mgf's are of that form.

Next your friend divides the resulting "U" by c and divide by Chi-Square with 100 degrees of freedom and he calls that "M." So then you can discover that what C is from that because you realize that U is VERY similar to M. Hint: If you know the distribution of U, then you can figure out what c is, and therefore, what distribution M is.

Lastly, B = $\frac{1}{M}$. So then B is, at least what I think, is the F-distribution with p = 100 and q = n+50, since you have two Chi-Squared distributions being divided by each other.

I hope this helps! Good Luck!

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