Suppose we have $Y_{i,n}, i \ge 1, n \ge 1 $ iid with expectation $\mu$. And given $Z_{n+1} = \sum_{i=1} ^{Z_n} Y_{i,n}$ and $Z_0 = 1$.

In lecture it was stated that $Y_{i,n}$ is independent of $Z_n$. I do not see how one would prove that. In particular we therefore used Wald's equation to obtain:

$E[Z_{n+1}] = E[Z_n] E[Y_{1,n}]$.

I would be glad if someone could explain to me why $Z_n$ is independent of $Y_{i,n}$ or if there is a different argument justifying $E[Z_{n+1}] = E[Z_n] E[Y_{1,n}]$.

Edit: The context was that $Z_n$ could be interpreted as the number of individuals in a generation and $Y_{i,n}$ as the number of children that individual $i$ in generation $n$ has.

  • $\begingroup$ Can you see that $Z_n$ is a function of $Y_{i,j}$ for $j < n$? (or more precisely, $Z_n$ is $\sigma(Y_{i,j} : i \geq 1, 1\leq j \leq n-1)$-measurable) $\endgroup$ Commented May 3 at 6:50
  • $\begingroup$ Yes, I see that $\endgroup$
    – user007
    Commented May 3 at 6:52
  • $\begingroup$ And can you see that $(Y_{i,n})_i$ is independent of $\sigma(Y_{i,j} : i \geq 1, 1 \leq j \leq n-1)$? $\endgroup$ Commented May 3 at 6:54
  • $\begingroup$ Is it independent since all $Y_{i,n}$ are independent? $\endgroup$
    – user007
    Commented May 3 at 6:57
  • 1
    $\begingroup$ I need to be a little careful with the subscripts, but essentially yes, since the entire 2d array $Y_{i,j}$ for $i \geq 1, j \geq 1$ are iid, the sigma algebras generated by any two disjoint subsets of this array will be independent. $\endgroup$ Commented May 3 at 7:01

1 Answer 1


First, note that $Z_n$ is $\sigma(Z_{n-1}, Y_{1,n-1}, Y_{2,n-1}, \ldots)$-measurable.

By induction on $n$, this shows that $Z_n$ is $\sigma(Y_{i,j} : i\geq 1, 1\leq j < n)$-measurable.

It now suffices to show that $\sigma(Y_{i,n} : i\geq 1)$ is independent of $\sigma(Y_{i,j} : i\geq 1, 1\leq j < n)$. However, this follows directly from the assumption that the 2D-array $(Y_{i,j})_{i\geq 1, \, j\geq 1}$ is an array of independent random variables, since the second subscript of the random variables generating $\sigma(Y_{i,n} : i\geq 1)$ is always $n$, which is larger than any of the second subscripts of the random variables generating $\sigma(Y_{i,j} : i\geq 1, 1\leq j < n)$.

  • $\begingroup$ Thank you very much! $\endgroup$
    – user007
    Commented May 3 at 7:20
  • $\begingroup$ do you happen to know a rigoruous proof for why the sigma-algebras are independ because the random variables are different and independent? $\endgroup$
    – user007
    Commented May 3 at 13:05

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