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Suppose $X_{0}, X_{1} ,....$ is a branching process who has an offspring distribution mean of $\mu$

Let

$$Y_{n}=\frac{X_{n}}{\mu^{n}}$$

I want to show that

$$E[Y_{n+1}|Y_{n}]=Y_{n}$$

Well,

I know that $E[X_{0}]=\mu$

and that $E[X_{n}]=\mu^{n}$

hence $E[Y_{n}]=E[Y_{n+1}]=1$

Now should I simply apply law of total expectation or some other basic? Or is there some key ideas I am missing. It is also possible I made mistakes in my reasoning above.

But I am really not sure, it seems that Y_{n} represents the actual size of the nth generation divided by the expected population of the nth generation

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    I suppose you can use the well known way of computing conditional expectation which involves calculating $E[Y_{n+1} | Y_n=y]$ and then in that expression replacing $y$ with $Y_n$. That way, by replacing the definition of $Yn$ you should end up with something involving $E[X_{n+1} | X_n=\mu^n y]$, which I guess you can calculate? – 2017-01-30
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    And correct me if I'm wrong, but $X_0=1$ in a branching process, so it's expectation is 1, not $\mu$ as you wrote. Perhaps you meant $E[X_1]=\mu$? – 2017-01-30
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    Yes that is what I meant, and write I am aware of that method but not sure how to apply it here – 2017-01-30
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    Can you calculate $E[X_{n+1} | X_n = x]$? I am not familiar with the branching process that much, but my guess would be $x\mu$, right? – 2017-01-30

1 Answers 1

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You can just plug the definition of $Y$ into $E[Y_{n+1} | Y_n = y]$, so you get $$E\left[\frac{X_{n+1}}{\mu^{n+1}} \left\vert \frac{X_n}{\mu^n}=y\right.\right]=\frac{1}{\mu^{n+1}}E\left[ X_{n+1}| X_n=\mu^n y\right].$$

Can you finish the rest?