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Math Help - Question of Variance

  1. #1
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    Question of Variance

    To be complete I'll just post everything about the problem.

    Q: Each brand of candy bar has one coupon in it. There are n different coupons in total; getting at least one coupon of each type entitles you to a prize. Each candy bar you eat can have any one of the coupons in it, with all being equally likely. Let X be the (random) number of candy bars you eat before you have all coupons. What are the mean and variance of X?

    A. So the expected value of X is a sum of geometric discrete random variables.

     X_1 = geom(1)
    so you have one coupon on the first trial.
    let X_2 denote the number of bars needed to get the 2nd coupon

     E[X_2] = geom(\frac{n-1}{n}) = \frac{n}{n-1}

     E[X_n] = geom(\frac{1}{n}) = n

     E[X] = E[\sum_{i=1}^{n} X_i] = \sum_{i=1}^{n} E[X_i]

    The part Im having trouble on is the variance. From my definition:
     Var(X) = E[(X-E[X])^2]
    so does that mean...

     Var(X) = E[(\sum_{i=1}{n} X_i - \sum_{i=1}{n}E[X_i])^2]

    Any help is appreciated
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  2. #2
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    Hello,

    There is an easier formula for the variance, derived from the one you have.

    Recall two properties of the expectancy :
    E[aX+Y]=aE[X]+E[Y] \quad (1) (where a is a real number)
    E[a]=a \quad (2) (again, where a is a real number)

    var(X)=E \bigg[(X-E[X])^2\bigg]

    var(X)=E \bigg[X^2+(E[X])^2-2XE[X]\bigg]

    By (1), we have var(X)=E[X^2]+E \left[(E[X])^2\right]-2 E[XE[X]]

    Note that E[X] is a value, a number. It is not a variable anymore.

    By (2), we have var(X)=E[X^2]+(E[X])^2-2 E[X]E[X]

    \boxed{var(X)=E[X^2]-(E[X])^2}


    Note that E[f(X)]=\sum_{x \in \chi} f(x) P(X=x)
    (here, f(x)=x and \chi is the set of all possible values for X)
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