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Math Help - Handling a moment generating function

  1. #1
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    Handling a moment generating function

    In my problem, I have the moment-generating function on a random variable X. The function is:
    M_{X}(t)=\frac{1}{(1-2500t)^4}

    I am to determine the standard deviation.

    I have no idea how to begin. I know that the standard form for a moment-generating function is:
    M_{X}(t)=E[e^{tx}]

    Neither my book nor Wikipedia give any clues as to what the variables mean. I certainly hope this is a situation where t (used in one explanation) = t (used in others). All I really know is, X is a random variable.

    Should I just be trying to transform what I am given into the E[e^{tx} form?
    Am I trying to determine what t is?
    Is the form of the function I was given an indication of a specific type of continuous distribution? None of the distributions I've seen (uniform, normal, etc.) has a moment-generating function that looks like that.
    Last edited by Boris B; April 19th 2008 at 07:40 PM.
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    Quote Originally Posted by Boris B View Post
    In my problem, I have the moment-generating function on a random variable X. The function is:
    M_{X}(t)=\frac{1}{(1-2500t)^4}

    I am to determine the standard deviation.

    I have no idea how to begin. I know that the standard form for a moment-generating function is:
    M_{X}(t)=E[e^{tx}]

    Neither my book nor Wikipedia give any clues as to what the variables mean. I certainly hope this is a situation where t (used in one explanation) = t (used in others). All I really know is, X is a random variable.

    Should I just be trying to transform what I am given into the E[e^{tx} form?
    Am I trying to determine what t is?
    Is the form of the function I was given an indication of a specific type of continuous distribution? None of the distributions I've seen (uniform, normal, etc.) has a moment-generating function that looks like that.
    The standard deviation is the (positive) square root of the variance. A formula for the variance is

    Var(X) = E(X^2) - [E(X)]^2.

    You can get these expected values from the moment generating function:

    Theorem: E(X^n) = \frac{d^n M_X(t)}{dt^n} \bigg{|}_{t=0}.

    Therefore:

    To get E(X), differentiate the moment generating function once and then substitute t = 0.

    To get E(X^2), differentiate the moment generating function twice and then substitute t = 0.
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    Quote Originally Posted by mr fantastic View Post
    The standard deviation is the (positive) square root of the variance. A formula for the variance is

    Var(X) = E(X^2) - [E(X)]^2.

    You can get these expected values from the moment generating function:

    Theorem: E(X^n) = \frac{d^n M_X(t)}{dt^n} \bigg{|}_{t=0}.

    Therefore:

    To get E(X), differentiate the moment generating function once and then substitute t = 0.

    To get E(X^2), differentiate the moment generating function twice and then substitute t = 0.
    Having re-read you original post, I think I should add the following:

    E(X) means the expected (or mean) value of the random variable X.
    E(X^2) means the expected (or mean) value of the random variable X^2.
    etc.
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    Hello,

    Nice ! I didn't know this way oO
    Though it's strange to me

    We learnt that E(X)=\frac{dM_X(t)}{dt} \bigg{|}_{t={\color{blue} 1}}

    E(X(X-1))=\frac{d^2 M_X(t)}{dt^2} \bigg{|}_{t=1}

    And so on...
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    Quote Originally Posted by Moo View Post
    Hello,

    Nice ! I didn't know this way oO
    Though it's strange to me

    We learnt that E(X)=\frac{dM_X(t)}{dt} \bigg{|}_{t={\color{blue} 1}}

    E(X(X-1))=\frac{d^2 M_X(t)}{dt^2} \bigg{|}_{t=1}

    And so on...
    Hi Moo. The standard normal distribution provides a simple counter-example to what you've learnt:

    The moment generating function is m(t) = e^{t^2/2} \Rightarrow m'(t) = t e^{t^2/2} \Rightarrow m'({\color{red}0}) = 0 = \mu.

    But m'({\color{red}1}) = \sqrt{e} \neq \mu ......
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    Is there a difference between the moment generating function of a random variable and the generating function of a random variable ?
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    Quote Originally Posted by Moo View Post
    Is there a difference between the moment generating function of a random variable and the generating function of a random variable ?
    The moment generating function of a random variable X is m = E(e^{tx}).

    What's the definition of the generating function of a random variable X?
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    Actually, I don't know if it's the same in French...

    What I call the generating function of a random variable is E(t^X)=\sum_x t^x P(X=x)

    where x represents all the possible values X can take...

    What does the moment generating function represent exactly ? In terms of probabilities I mean
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    Quote Originally Posted by Moo View Post
    Actually, I don't know if it's the same in French...

    What I call the generating function of a random variable is E(t^X)=\sum_x t^x P(X=x)

    where x represents all the possible values X can take...

    What does the moment generating function represent exactly ? In terms of probabilities I mean
    Moment-generating function - Wikipedia, the free encyclopedia

    Your generating function does not look too easy to calculate for continuous distributions .....? Do you know off-hand what it is for the standard normal distribution (I'm too lazy to calculate it)?
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  10. #10
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    I will have difficulties to understand since there isn't the French page on wikipedia :/
    But now I see where the difference is : mine is not for continuous distributions.
    We'd use it for Poisson, Binomial, Geometric, etc... laws
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  11. #11
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    I wonder if the two can be linked... For example if we know what the distribution for X is, it'd be possible to retrieve one from another.

    What I still don't understand is this difference t=1 or t=0. Maybe it has to deal with a substitution when transforming P(X=x) into an integral...

    Oh nevermind...I can't think about it right now
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  12. #12
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    Quote Originally Posted by Moo View Post
    I wonder if the two can be linked... For example if we know what the distribution for X is, it'd be possible to retrieve one from another.

    What I still don't understand is this difference t=1 or t=0. Maybe it has to deal with a substitution when transforming P(X=x) into an integral...

    Oh nevermind...I can't think about it right now
    There's no big mystery.

    The generating function g(t) is easily calculated for discrete probability distributions. For example:

    For the geometric distribution:

    g(t) = \sum_{x = 1}^{\infty} t^x (1 - p)^{x - 1} p = p t \sum_{x = 1}^{\infty}  [t (1 - p)]^{x - 1} = \frac{pt}{1 - t(1 - p)}.

    g'(t) = \frac{p}{(t(p - 1) + 1)^2} \Rightarrow g'(1) = \frac{p}{p^2} = \frac{1}{p} = \mu.

    It's well known that the moment generating function is m(t) = \frac{p e^t}{1 - (1 - p)e^t}.

    m'(t) = \frac{p e^t}{e^t (p - 1) + 1)^2} \Rightarrow m'(0) = \frac{p}{p^2} = \frac{1}{p} = \mu.

    So g'(1) = m'(0) = \mu - no big deal.


    In fact, it's not hard to see that if g(t) = \sum_x t^x f(x), then g'(1) = \sum_x x f(x) = E(X).
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  13. #13
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    I really have to read your link about the moment generating function, thanks again for it
    Actually, I'm far too tired to think properly
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