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Let $X_0, X_1, ...$ be a branching process with offspring distribution mean $\mu$. Let $Y_n = \frac{X_n}{\mu^n}, n \ne 0$. Show $E[Y_{n+1} | Y_n] = Y_n$

My attempt:

$E[Y_n] = E[\frac{X_n}{\mu^n}] = \frac{1}{\mu^n}E[X_n] = \frac{1}{\mu^n}\mu^n = 1$

But I'm not sure how to continue

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    What is the random variable $Z_n$? – 2017-02-10
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    @user375366 sorry it should be $X_n$ – 2017-02-10
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    You should try to calculate $E[Y_{n}|Y_{n-1}]$, not $E[Y_n]$. – 2017-02-10

1 Answers 1

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Let $Z_i$ denote the number of offspring coming from the $i^{\text{th}}$ member.

$ \begin{align*} E(Y_{n+1}\mid Y_n)&=E(\frac{X_{n+1}}{\mu^{n+1}}\mid\frac{X_n}{\mu^n}) \\ &=\frac{1}{\mu^{n+1}}E(X_{n+1}\mid X_n) \\ &=\frac{1}{\mu^{n+1}}E(\sum_{i=1}^{X_n}Z_i\mid X_n) \\ &=\frac{1}{\mu^{n+1}}X_n\mu \\ &=\frac{X_n}{\mu^n} \\ &=Y_n. \end{align*} $

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    Why did you compute $E(Y_n | Y_{n-1})$ instead of $E(Y_{n+1} | Y_n)$? – 2017-02-10
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    @user2980566 Because I misread the question. – 2017-02-10