# Probability Theory watch

1. 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)
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.
3. (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
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.
5. (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
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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Updated: March 2, 2018
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