MHB Probability distribution of a RV is a function of another RV

[bincy]

Dear friends,

I have a Random Variable I. Sample space of I is from 1,2,3... inf(countably infinite). It's probability distribution P(I=i) is a function of another set of Random Variables Xi's, which are uniformly distributed in [0,1]. These Random variable are iid. I have to find out the mean of I. The mean of Xi's is 0.5 for every i.

I am giving the probability distribution of i below.

[math]P(I=i)=\left\{ \prod_{j=1}^{i-1}\left(1-\left(N*x_{j}*\left(1-x_{j}\right)^{N-1}\right)\right)\right\} N*x_{i}\left(1-x_{i}\right)^{N-1} [/math]

The average of I is [math]\sum_{i=1}^{inf}P(I=i)*i [/math]

Can anyone give any idea to solve this problem?regards,
Bincy

---------- Post added at 16:41 ---------- Previous post was at 15:13 ----------

[math] \int_{0}^{1}\int_{0}^{1}.....\int_{0}^{1}\left\{ \prod_{j=1}^{i-1}\left(1-\left(N*x_{j}*\left(1-x_{j}\right)^{N-1}\right)\right)\right\} N*x_{i}\left(1-x_{i}\right)^{N-1} *i [/math]

Is it the actual average? If so, how to simplify it? Here integration is infinite times.regards,
Bincy

---------- Post added at 17:44 ---------- Previous post was at 16:41 ----------

I would like to add one more general point and kindly seeking the confirmation from you.

That is,

If the prob distribution of a RV X is a fn of another RV Y, X itself would be a fn of Y.

 

[bincy]

The answer(mean of I) is N since the probability of success in the 1st, 2nd, 3rd and so on is geometric with parameter 1/N. Ignore my 2nd and 3rd posts