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Math Help - Find Mean/Variance from probability mass function

  1. #1
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    Find Mean/Variance from probability mass function

    I'm not sure if this actually Advanced, but it is from a 3000-level University course. My apologies if I am misposting.

    A random variable X has the probability mass function
    f(x) = p * 0.6^{x-1} , for x = 1, 2 , 3,
    Find p, the mean and the variance of X.
    Ok... So that seems pretty clear to me that p = .4 (based on the logic that q=.6 = 1-p).

    But I have no idea where to go from here. If I go to take the mean and variance, according to the text and notes, the formulae I'd be using are:

    <br />
\mu = \sum x f(x) = \sum_{x=1}^{\infty} x (.4 * .6^{x-1})<br />

    and

    <br />
\delta^2 = \sum(x-\mu)^2f(x)<br />

    Maybe my calculus-fu has weakened (it's been a long time), but I can't for the life of me figure out how to do those sums since they don't appear to resemble any of the series I studied in Calc II...

    If anyone could give me a pointer, or tell me where I did something wrong, I'd appreciate it. Thanks...
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  2. #2
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    Quote Originally Posted by komissar View Post
    I'm not sure if this actually Advanced, but it is from a 3000-level University course. My apologies if I am misposting.

    Ok... So that seems pretty clear to me that p = .4 (based on the logic that q=.6 = 1-p).
    That is the right value for p, but your explanation cannot be followed. You should state that for f to be a pmf:

    \sum_{x=1}^{\infty} p (0.6)^{x-1}=1

    or state some other obvious reason why this is so.

    CB
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  3. #3
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    Quote Originally Posted by komissar View Post
    But I have no idea where to go from here. If I go to take the mean and variance, according to the text and notes, the formulae I'd be using are:

    <br />
\mu = \sum x f(x) = \sum_{x=1}^{\infty} x (.4 * .6^{x-1})<br />
    Consider:

    g(x)=0.4 \; \sum_{x=1}^{\infty} 0.6^{x}

    Now differentiate g(x) (and the series term by term)

    CB
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  4. #4
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    Quote Originally Posted by komissar View Post
    ...
    and

    <br />
\sigma^2 = \sum(x-\mu)^2f(x)<br />

    Use:

    \sigma^2= \overline{x^2}-\overline{x}^2 = \overline{x^2} - \mu^2

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    That is the right value for p, but your explanation cannot be followed. You should state that for f to be a pmf:

    \sum_{x=1}^{\infty} p (0.6)^{x-1}=1

    or state some other obvious reason why this is so.
    I was just putting that there to illustrate my thought process in case that's where my error was, but I'll keep it in mind. Thank you.


    Quote Originally Posted by CaptainBlack View Post
    Consider:

    g(x)=0.4 \; \sum_{x=1}^{\infty} 0.6^{x}

    Now differentiate g(x) (and the series term by term)
    I'm a little lost there. Is that a rewriting of the function I had, or was mine wrong?



    Quote Originally Posted by CaptainBlack View Post
    Use:

    \sigma^2= \overline{x^2}-\overline{x}^2 = \overline{x^2} - \mu^2

    CB
    I assume it'll be summed the same way you described for the mean above?

    Thanks again.
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