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    I'm doing exponential distribution. A call costs 15p per minute or part of. Cost of a call is X pence. T minutes is the duration with pdf
    f(t)= ae^(-at).

    I need to prove that E(X) = 15/(1-e^(-a))

    I already know P(X=15r) = e^(-ra).(e^(a)-1)
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    (Original post by Jaden1998)
    I'm doing exponential distribution. A call costs 15p per minute or part of. Cost of a call is X pence. T minutes is the duration with pdf
    f(t)= ae^(-at).

    I need to prove that E(X) = 15/(1-e^(-a))

    I already know P(X=15r) = e^(-ra).(e^(a)-1)
    You just then need to sum 15re^(-ra).(e^(a)-1) to find the expectation.
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    (Original post by RichE)
    You just then need to sum 15re^(-ra).(e^(a)-1) to find the expectation.
    Are you able to go into more detail about the layout and how to start that, I'm still a bit confused. Thank you
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    (Original post by Jaden1998)
    Are you able to go into more detail about the layout and how to start that, I'm still a bit confused. Thank you
    The sum of kx^k can be worked out by differentiating the sum of x^k and the latter is a geometric series.
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    (Original post by RichE)
    The sum of kx^k can be worked out by differentiating the sum of x^k and the latter is a geometric series.
    So I calculated my first term as 1 - e^(-a) and my common ratio as e^(-a) but when I sub unto geometric series formula I get (1 - e^(-a))^15 which is not what I want
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    (Original post by Jaden1998)
    So I calculated my first term as 1 - e^(-a) and my common ratio as e^(-a) but when I sub unto geometric series formula I get (1 - e^(-a))^15 which is not what I want
    As I said above the series isn't a geometric series, but it is the derivative of one.

    As an alternative perhaps you know the binomial series for (1-x)^(-2).
 
 
 

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