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Math Help - Help finding the expectations with the PMF's

  1. #1
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    Help finding the expectations with the PMF's




    Hey, I'm having trouble understanding the process to find the expectation of pmf's that I have to find.

    I know the formula for it but still cannot understand it.

    With The first link, I can get all the way to question (c)

    I get the pmf, P(X=x) = 1/4*(2/3)^{k-2}    for (K=1,2,3,.....)

    But can't get the expectation, and for the second link i'm completely clueless
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  2. #2
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    That formula isn't quite right (plug in the rat getting it on the first try). It seems you have everything, but the fact that he chooses correct on the current trial. For example, suppose the rat chooses correctly on the 2nd try. By your formula the rat has a .25 chance of that happening, which actually refers to him choosing correctly on the first try. So it should be:

    Pr(X=x)=\left(\frac{1}{4}\right)\left(\frac{2}{3}\  right)^{x-2}\left(\frac{1}{3}\right) , for x=2,3. . .

    Where the 1/3rd refers to him getting it on the xth trial beyond the first.

    The expectation for this would be done as you do any expectation - the sum of of x times the probability of x. However - the actual calculation is probably going to be super tedious. It simplifies down, but you are having to do a kinda crummy summation.

    The same goes for your second problem. You know the summation of all those probabilites have to add up to 1; so using the definition you're going to have to use some series trickery to figure out what A is, as well as the expecation.

    I'll maybe work on it later if you haven't gotten it down yet. Good luck!
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  3. #3
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    Thanks for correcting me on that one, slight oversight on my part!

    The calculation of finding the expectation is the part i'm struggling with.

    I know the formula for it in my mind, except I don't really know how to do it, because this PMF we have can take x > 2, it doesn't stop, so How do I do it? Sorry if this is basic my head just isn't wrappin round it.

    You don't have to do the calculation for me, just a little hint on the above would help

    Thank you for the help
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  4. #4
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    I wouldn't say its basic. Just because your trials could (in theory) go on forever, doesn't mean ON AVERAGE it would (easiest example, Let X be the number of coin flips until you get a heads - in theory, you could flip to 1million - not likely though). In the summation notation you are going to go from 2 to infnity, but more than likely the summation (well not more than likely. . .it must) has a limit. The summation I came up with (after cleaning it up):

    E(X)=\frac{1}{4}+\frac{9}{16}\sum x\left( \frac{2}{3} \right)^{x}

    What that is, I have no idea - you might have use some results from Calc II and series approximations. Blargh.
    Last edited by ANDS!; January 20th 2011 at 05:39 PM. Reason: Corrected Formula
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  5. #5
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    I've just had a check of your PMF you suggested, but realised that it doesn't give you the right probabilities,

    P(X=1) = \frac{1}{4}
    P(X=2) = \frac{3}{4} * \frac{1}{3} = \frac{1}{4}
    P(X=3) = \frac{3}{4} * \frac{2}{3} * \frac{1}{3} = \frac{1}{5}

    Where as with P(X=2) on your summation it doesn't give 1/4.

    Mine seems to work for all K = 2,3,4 etc

    I've checked the solutions and they seem to concur with the Summation I had...

    Here's what they did for the Expectation

    E(X)=\frac{1}{4}+\frac{1}{4}(\sum k\left\frac{2}{3}^{k-2})

    E(X)=\frac{1}{4}+\frac{3}{8}(\sum k(\frac{2}{3})^{k-1}- 1)

    E(X)=\frac{1}{4}+\frac{3}{8}((1-\frac{2}{3})^{-2)}-1)

    E(X)=\frac{1}{4} + \frac{3}{8}(9-1) = \frac{13}{4}

    I just don't understand what happened from Line 2 to Line 3, I've looked at it for quite a while and no idea
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  6. #6
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    My bad, it should be 3/4 not 1/4 in the formula I posted. The reason yours "works" is a bit of kismet action: 3/4*1/3 (which was missing) is 1/4. As for whats going on in the summation:

    From the second line to the third, all they have done is adjusted the summation, so that it works for ALL values of "k" starting at 1: Looks like the first pulled off a \left( \frac{2}{3} \right)^{-1} to get k\left( \frac{2}{3} \right)^{k-1}. This summation is only valid for starting at 2, so they adjusted the index to start at 1, and subtract 1 from the total series to account for that. From there I guess they used properties of series to simplify - would have to do some digging to see how.
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