0

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 $$E[3N_5^4 + N_8^3 | N_0, N_1, N_2] = E[3N_5^4 + N_8^3 | N_2]$$

Vahan
  • 415
  • 2
  • 14
  • 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. – Ian May 08 '19 at 12:53
  • 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). – Ian May 08 '19 at 12:56
  • Then what happens to the conditional? And where are you getting $E[X(X-1)...]$ Are you using a generating function? – Vahan May 08 '19 at 13:37
  • 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. – Ian May 08 '19 at 14:00
  • 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? – Vahan May 08 '19 at 18:39
  • It's the same thing as what I suggested, just written out in more detail. – Ian May 08 '19 at 18:41
  • So would you express the problem $$E(N_2 + 3N_5^4 + N_8^3 - N_2 | N_2)?$$ – Vahan May 08 '19 at 19:15
  • No, you have to add-and-subtract inside the exponents, so for example $E[N_5^4 \mid N_2=n_2]=E[(n_2+N_3)^4]$. – Ian May 08 '19 at 19:18
  • I think I found what you mean by the products. To find $E(N_n^2)$, use the $X_n$: $$E(N_n^2) = \sum_{i=1}^n E(X_i^2) + \sum_{i=1}^n \sum X_i X_j$$ where i not equal j. The result is $$np + n(n-1)p^2 = n^2p^2 + npq$$ I fail to see how this works. For instance consider $E(N_2^2)$. $$(X_1 + X_2)^2 = X_1^2 + 2X_1X_2 + X_2^2$$ The expected value of this is $2p^2 + 2p$. But the above formula would give $4p^2 + 2pq$. Also I'm not sure how to generalize the method for expectation to higher exponents. And it seems there should be a notation for coefficients. – Vahan May 08 '19 at 20:46
  • That's not what I'm talking about. Say you want to compute $E[X^3]$ for the binomial(n,p) distribution. You can do that this way: instead look at $E[X(X-1)(X-2)]=\sum_{i=3}^n i(i-1)(i-2) {n \choose i} p^i q^{n-i}$, then write $i(i-1)(i-2) {n \choose i}=\frac{n!}{(n-i)!} \frac{i(i-1)(i-2)}{i!}=\frac{n!}{(n-i)!(i-3)!}$. – Ian May 08 '19 at 22:07
  • This looks a lot like ${n-3 \choose i-3}$; in particular, the ratio of these two numbers doesn't depend on $i$. So now you can factor stuff out of the sum until the summand looks like ${n-3 \choose i-3} p^{i-3} q^{(n-3)-(i-3)}$ and then use that the binomial PMF sums to 1 in order to finish out the calculation. But now $E[X(X-1)(X-2)]$ is some combination of $E[X],E[X^2]$ and $E[X^3]$ so if you know two of them then you can derive the other one. Of course you could equally well achieve the same goal by using the PGF instead. – Ian May 08 '19 at 22:08
  • Could you make use of the multinomial expansion? – Vahan May 09 '19 at 18:28
  • I actually didn't need to calculate that, please see my edit. – Vahan May 10 '19 at 01:52

0 Answers0