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Thread: Time scaling a geometric random variable

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    Time scaling a geometric random variable




    Hello,

    When one performs bernoulli trials at discrete times t = 1, 2, 3, ....., with the probability of a success at an arbitrary trial given by p, then the geometric random variable represents the number of trials required to achieve one success.

    Please help me confirm the correct answer to this problem:
    Given a geometric random process where the bernoulli trials are performed only every Nth time step, i.e. t = N, 2N, 3N, ..... infinity.

    In my opinion, the latter scenario is the same as the former scenario except that the latter scenario has a mean time = N / p. Is this correct?

    In the latter scenario, how does the distribution look like, what is its mean and standard deviation.

    Thank you very much.

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    Re: Time scaling a geometric random variable

    Well if we just let X ~ Geo(p) then X can represent the outcome from your first paragraph. Let Y be a random variable associated with the process you're interested in. It's easy to see that Y = N*X. So we can just use facts like if Z is a random variable then E[c*Z] = c*E[Z] for any real value of c and SD[c*Z] = sqrt(c)*SD[Z]
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    Re: Time scaling a geometric random variable

    Thanks Dason. I thought so too. I am trying to formulate the problem as:
    Y ~ Geom(p/N). Is this correct?

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    Re: Time scaling a geometric random variable

    Not really - Y itself won't be geometric. It's a scaled geometric but you can't call it a geometric random variable unless you just redefine your time units (and I don't really see a reason to not do that for this problem)
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    Re: Time scaling a geometric random variable

    Yes, I was attempting that. Now, lets get into the details. In order to compute the mean a geom. r.v.:
    sum^{Infinity}_{x = 1} x (1-p)^(x-1) p

    Performing the time translation,

    sum^{Infinity}_{x = 1} N x (1-p)^(N x-1) p

    Let y = x N
    Then,

    sum^{Infinity}_{y = N} y (1-p)^(y-1) p

    Now, this can be solved as

    sum^{Infinity}_{y = 1} y (1-p)^(y-1) p
    - sum^{N-1}_{y = 1} y (1-p)^(y-1) p

    The summations can be solved with the help of the online tool:
    http://www.wolframalpha.com/widget/w...showWarnings=1


    However, the resulting mean seems to be less than 1 which is not possible since the minimum possible mean time = 1. Let me know if this approach matches with yours.

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    Re: Time scaling a geometric random variable

    It's hard to read your math - take a look at this: http://www.talkstats.com/showthread....-use-Math-tags

    I think your second equation is wrong though. But why exactly are you trying to do this? The mean of a geometric is already known and yours is just a linear transformation. The mean of Y is N/p
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    Re: Time scaling a geometric random variable

    Excuse my math presentation. My objective is to derive the parameters (mean, std dev) using first principles when time scaling a geom. r.v and i came across the above mentioned issues. The following paper tackles this issue but i found it very counter intuitive:
    http://www.advancesindifferenceequat...7-2012-211.pdf
    You may jump to page 9, equation 4.6 and 4.7.

    In fact, the paper contradicts your notion of the mean and std dev when considering time scaling of geom. r.v.

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    Re: Time scaling a geometric random variable

    Where is the contradiction? They say that the mean for a geometric is 1/p. I say that as well. But then once you 'time scale' it you need to account for that - which is where the N/p comes from...
    I don't have emotions and sometimes that makes me very sad.

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    Re: Time scaling a geometric random variable

    Let ph = p/N. Then, the mean as you say is 1/ph = N/p.

    The main contradiction occurs with the expression for the variance. Now, substitute ph = p/N in equation 4.7. The variance is no longer sqrt(c)*SD[Z] as you mentioned in your post.

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    Re: Time scaling a geometric random variable

    You're right I do have a mistake - the formula should be SD[c*Z] = c*SD[Z]. I was thinking of sums of iid random variables - my apologies.

    Note however that standard deviation isn't variance - be careful with your wording.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Time scaling a geometric random variable

    Thanks. I am aware that std dev = sqrt(Variance).

    Going by what you mentioned, std = N * sqrt( \frac{1 - p}{p^{2}} ).

    However, by substituting ph = p/N in equation 4.7 in the paper, the
    std dev = sqrt(N) * sqrt( \frac{N - p}{p^{2}} ), which is not equal to the above equation.

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    Re: Time scaling a geometric random variable

    Quote Originally Posted by mathRookie2013 View Post
    However, by substituting ph = p/N in equation 4.7 in the paper
    Why are you doing this?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Time scaling a geometric random variable

    Lets take a step back.
    p = prob. of success parameter of the original unscaled geom. r.v.
    In the paper, ph = prob. of success parameter of the scaled geom. r.v.
    The paper says that the mean of the scaled r.v. is given by 1 / ph. On the other hand, you say that the mean should be N / p. Both statements can be true only if ph = p / N.

    If we agree on the above argument, lets set ph = p / N. As per the paper, the std. dev. of the scaled r.v. when setting ph = p / N is given by
    std dev = sqrt(N) * sqrt( \frac{N - p}{p^{2}} ).

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    Re: Time scaling a geometric random variable

    The paper is doing some else - it's looking at the expected number of 'trial groups' one would need to run to get a success. Since each trial group has 'h' events in it (which is the same was what we're referring to as N) there is no contradiction.

    If you want to know the expected time on the original scale it is N/p. If you want to know the expected number of trial groups that will be run it is 1/p.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Time scaling a geometric random variable


    Thanks Dason. Things are clear now. Thanks for your patience throughout this discussion.

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