Expectation with independant random variables

Question

Suppose we have a sequence of functions $\{f^1(\cdot),...,f^m(\cdot)\}$ and i.i.d random variables $\{i_k\}_{k\geq 1}$ drawn with replacement from the set $\{1,...,m\}$. Suppose at each $k$, $x_k$ is dependant on $i_k$ (where $x_k$ is some random variable generated after realising $i_k$). Also, $i_k$ is independent of $x_{k-1}$. Can we claim that :

$$ \mathbb{E}[f^{i_k}(x_{k-1}) ] = \mathbb{E}[f^{i_{k-1}}(x_{k-1}) ] $$

My intuition is that since $i_k$ are i.i.d, the above claim is true. But the fact that is bothering me is that $i_{k-1}$ and $x_{k-1}$ are dependant.

I am sorry if the question is trivial or does not make sense. Thanks in advance.

Answer 1

Suppose $m=2$, $x_k=i_k$ and $f^1(x)=x$ and $f^2(x)=2x$

Then with $k=2$ you have

• $f^{i_k}(x_{k-1}) = 1$ when $i_1=1$ and $i_2=1$
• $f^{i_k}(x_{k-1}) = 2$ when $i_1=1$ and $i_2=2$
• $f^{i_k}(x_{k-1}) = 2$ when $i_1=2$ and $i_2=1$
• $f^{i_k}(x_{k-1}) = 4$ when $i_1=2$ and $i_2=2$

making $\mathbb{E}[f^{i_k}(x_{k-1}) ] = 2.25$

but

• $f^{i_{k-1}}(x_{k-1}) =1$ when $i_1=1$
• $f^{i_{k-1}}(x_{k-1}) =4$ when $i_1=2$

making $\mathbb{E}[f^{i_{k-1}}(x_{k-1}) ] =2.5$ and this is different