I just want to make sure I'm understanding conditional expectation correctly:

Let $X_1,X_2,X_3$ denote three independent coin flips with probability of heads $\frac{1}{4}$ and probability of tails $\frac{3}{4}$, and $X_i=2$ if heads and $X_i=0$ if tails.

Then I'm looking to determine the conditional probability


where $F_1$ is the sigma algebra generated by $X_1$, or equivalently that generated by the partition $\{\varnothing,\Omega,\{HHH,HHT,HTH,HTT\},\{THH,THT,TTH,TTT\}\}$.

So I compute this by taking $P(\{HHH,HHT,HTH,HTT\})=\frac{1}{64}+\frac{3}{64}+\frac{3}{64}+\frac{9}{64}=\frac{1}{4}$

and then weighting these according to the various sums: $6\frac{1}{64}+4\frac{3}{64}+4\frac{3}{64}+2\frac{9}{64}=\frac{3}{4}.$

And then doing the same thing for the other set



And then taking $\frac{\frac{3}{4}}{\frac{1}{4}}=3$ and $\frac{\frac{3}{4}}{\frac{3}{4}}=1$, to obtain the random variable:

$$\mathbb{E}[X_1+X_2+X_3|F_1](\omega)=3 \;for\; \omega\in\{HHH,HHT,HTH,HTT\}$$ $$\mathbb{E}[X_1+X_2+X_3|F_1](\omega)=1 \;for\; \omega\in\{THH,THT,TTH,TTT\}.$$

Can anyone please tell me is this correct? Thanks.


More quickly, and for the same result, note the following:

  1. The random variable $X_1$ is $F_1$-measurable hence $E[X_1\mid F_1]$.
  2. For every $k\ne1$, the random variable $X_k$ is independent of $F_1$ hence $E[X_k\mid F_1]=E[X_k]$.
  3. For every $k$, $E[X_k]=\frac12$.

Thus, $$E[X_1+X_2+X_3\mid F_1]=X_1+\tfrac12+\tfrac12=X_1+1$$

  • $\begingroup$ Ok cool yah you're right, although I think you mean $E[X_k]=\frac{1}{2}$, since the coin is biased. $\endgroup$ – Thoth Jun 4 '13 at 22:01

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