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Math Help - Mathematical expectation of r. v.

  1. #1
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    Mathematical expectation of r. v.

    Hi members,
    A coin is tossed until head appears. What are the expected number of tosses ?
    Answer : Two tosses. What would be your answer ?
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  2. #2
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    Re: Mathematical expectation of r. v.

    $Pr[k]=\left(\dfrac 1 2 \right)^k$

    $E[K]=\displaystyle{\sum_{k=1}^\infty}k \left(\dfrac 1 2 \right)^k= 2$
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  3. #3
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    Re: Mathematical expectation of r. v.

    Essentially the same as romsek. The probability a head occurs on the first toss is 1/2. In order that the first head occur on the second toss, there must be a tail on the first toss and a head on the second. The probability of each of those is 1/2 so the probability the first head occurs on the second toss is 1/4. Generally the probability the first head occurs on the kth toss is the probability of a head times the probability the first k-1 tosses are tails: \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{  k-1}= \left(\frac{1}{2}\right)^k as romsek said. The "Mathematical expectation" is the value of a given outcome times its probability, summed over all outcomes: \sum_{k= 1}^\infty k\left(\frac{1}{2}\right)^k, again what romsek gave.

    One way to see that the sum is, in fact, 2 is to notice that the derivative of x^k is kx^{k- 1}= x^{-1}(kx^k). If we were to differentiate \sum_{k= 0}^\infty x^k term by term, we would get \sum_{k=1}^\infty kx^{k- 1}= x^{-1}\sum_{k= 1}^\infty kx^k. That first sum is, of course, a geometric series which has sum \frac{1}{1- x}. The derivative of that is \frac{1}{(1- x)^2}. Now, if x= 1/2, that is \frac{1}{(1- 1/2)^2}= 4. So we are saying that (1/2)^{-1}\sum_{k=1}^\infty k(1/2)^k= 2\sum_{k=1}^\infty k(1/2)^k= 4 so that \sum_{k= 1}^\infty k(1/2)^k= 2.
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  4. #4
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    Re: Mathematical expectation of r. v.

    Essentially the same as romsek. The probability a head occurs on the first toss is 1/2. In order that the first head occur on the second toss, there must be a tail on the first toss and a head on the second. The probability of each of those is 1/2 so the probability the first head occurs on the second toss is 1/4. Generally the probability the first head occurs on the kth toss is the probability of a head times the probability the first k-1 tosses are tails: \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{  k-1}= \left(\frac{1}{2}\right)^k as romsek said. The "Mathematical expectation" is the value of a given outcome times its probability, summed over all outcomes: \sum_{k= 1}^\infty k\left(\frac{1}{2}\right)^k, again what romsek gave.

    One way to see that the sum is, in fact, 2 is to notice that the derivative of x^k is kx^{k- 1}= x^{-1}(kx^k). If we were to differentiate \sum_{k= 0}^\infty x^k term by term, we would get \sum_{k=1}^\infty kx^{k- 1}= x^{-1}\sum_{k= 1}^\infty kx^k. That first sum is, of course, a geometric series which has sum \frac{1}{1- x}. The derivative of that is \frac{1}{(1- x)^2}. Now, if x= 1/2, that is \frac{1}{(1- 1/2)^2}= 4. So we are saying that (1/2)^{-1}\sum_{k=1}^\infty k(1/2)^k= 2\sum_{k=1}^\infty k(1/2)^k= 4 so that \sum_{k= 1}^\infty k(1/2)^k= 2.
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