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Consider a Bernoulli process where at each step or time there is a success, 1, or failure, 0. $N_k$ denotes the number of successes at the $k^{th}$ step. The probability of success is $p$ and the probability of failure is $q$, where $p + q = 1$.

The problem is to find $$E[3N_5^4 + N_8^3 | N_0, N_1, N_2]$$

I don't understand the problem. I thought that past N have no influence on future N.

I know that $E(N_n) = np$ and $E(N_n^2) = n^2p^2 + npq$, but I don't know how to tackle those greater exponentials.

EDIT:

The problem was actually to show that $$E[3N_5^4 + N_8^3 | N_0, N_1, N_2] = E[3N_5^4 + N_8^3 | N_2]$$

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  • $\begingroup$ Past $N$ absolutely have an influence on future $N$ because it's just the sum of all the successes up to that point. But the $N$ before $N_2$ are redundant, because the future $N$ don't care when the successes up to $k=2$ were, just how many of them there were. $\endgroup$ – Ian May 8 at 12:53
  • $\begingroup$ Anyway, a place to start is to remove the conditional by writing $N_5=n_2+N_3',N_8=n_2+N_6'$ where the $N'$ have the same distribution as the $N$ (but they are obtained by summing a different subset of Bernoulli variables). Then after some algebra you reduce things to finding the higher moments of the binomial distribution, which you can do the same way you found the first two (start by finding $E[X(X-1)(X-2)]$ and $E[X(X-1)(X-2)(X-3)]$ and then put the pieces together the way you want). $\endgroup$ – Ian May 8 at 12:56
  • $\begingroup$ Then what happens to the conditional? And where are you getting $E[X(X-1)...]$ Are you using a generating function? $\endgroup$ – Vahan May 8 at 13:37
  • $\begingroup$ The conditional only appears through the known value of $N_2$; for example $N_5=N_2+X_3+X_4+X_5$ and now you can replace $N_2=n_2$. That product trick is just a way to compute higher moments of the binomial distribution by exploiting properties of factorials. You probably saw it already in the derivation of the variance. $\endgroup$ – Ian May 8 at 14:00
  • $\begingroup$ After replacing, does that eliminate the conditional |? I think I've only seen that product with use of a generating function $E(\alpha^x)$ where the derivatives evaluated at one give those products. There is a trick in the book for conditionals. For example, find $E(N_5 | N_2)$. This can be expressed as $E(N_2 + (N_5 - N_2) | N_2)$, which is $$E(N_2 | N_2) + E(N_5 - N_2 | N_2) = N_2 + 3p$$ Is this of use here? $\endgroup$ – Vahan May 8 at 18:39

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