$E(X)$ in recursive relations

Question

If $$X_{n} = (1.6)X_{n-1}$$ for probability 0.4 and $$(0.7)X_{n-1}$$ for probability 0.6, how can I get $E(X_{n})$? ($X_0>0$)

I tried $$E(X_{n})= E(X_{n-1})((1.6)(0.4)+(0.7)(0.6))$$ but I think it is not logically correct.

Answer 1

You could say $X_n=(1.6)^Q(0.7)^{n-Q}X_0$ where $Q\sim \text{Binomial}(n,0.4)$. Then $$\mathbb{E}(X_n)=X_0\sum_{q=0}^n \left(1.6\right)^{q}\left(0.7\right)^{n-q}\mathbb{E}(Q=q)$$ Another approach is by using total law of expectation: $$\mathbb{E}(X_n)=\mathbb{E}\left(\mathbb{E}(X_n|X_{n-1})\right)=\mathbb{E}(X_{n-1})\left[\left(1.6\right)\left(0.4\right)+\left(0.7\right)\left(0.6\right)\right]$$ Note this is the recursive relation you obtain. Using either approach you'll get $$\mathbb{E}(X_n)=X_0\left(\frac{53}{50}\right)^{n}$$

Answer 2

Independence is a crucial assumption.

For instance, suppose we start with $X_0 = 1$ and also fix a random variable $R$ such that $R = 1.6$ with probability $0.4$ and $R=0.7$ with probability $0.6$.

Then $X_n = R^n$ satisfies all your given assumptions, but your recursive equation for $E[X_n]$ fails. To be explicit: $$E[X_n] = 1.6^n(0.4) + 0.7^n(0.6)$$ while $$\begin{align*}E[X_{n-1}]&((1.6)(0.4) + (0.7)(0.6)) \\ &= 1.6^n(0.4)^2+1.6(0.7)^{n-1}(0.4)(0.6)+1.6^{n-1}(0.7)(0.4)(0.6)+0.7^n(0.6)^2.\end{align*}$$ You may check that these are not equal.

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On the other hand, if we assume that $\{X_n = 1.6X_{n-1}\}$ and $\{X_n=0.7X_{n-1}\}$ are independent of $X_{n-1}$, then this fixes the above issue. Writing $A_n = \{X_n = 1.6X_{n-1}\}$ and using the law of total expectation, we have

$$\begin{align*}E[X_n] &= E\Big[E[X_n\mid X_{n-1}]\Big] \\ &= E\Big[E[1.6X_{n-1}1_{A_n}+0.7X_{n-1}1_{A_n^c}\mid X_{n-1}]\Big] \\ &=E\Big[1.6X_{n-1}E[1_{A_n}\mid X_{n-1}]+0.7X_{n-1}E[1_{A_n^c}\mid X_{n-1}]\Big] \\ \textit{(by indep.)}&= E\Big[1.6X_{n-1}E[1_{A_n}]+0.7X_{n-1}E[1_{A_n^c}]\Big] \\ &= E[X_{n-1}](1.6E[1_{A_n}]+0.7E[1_{A_n^c}]) \\ &= E[X_{n-1}]((1.6)(0.4)+(0.7)(0.6))\end{align*}$$