# Understanding Conditional Expectation

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

$$\mathbb{E}[X_1+X_2+X_3|F_1],$$

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

$P(\{THH,THT,TTH,TTT\})=\frac{3}{64}+\frac{9}{64}+\frac{9}{64}+\frac{27}{64}=\frac{3}{4}$

$4\frac{3}{64}+2\frac{9}{64}+2\frac{9}{64}+0\frac{27}{64}=\frac{3}{4}.$

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.

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, $$E[X_1+X_2+X_3\mid F_1]=X_1+\tfrac12+\tfrac12=X_1+1$$
• Ok cool yah you're right, although I think you mean $E[X_k]=\frac{1}{2}$, since the coin is biased. – Thoth Jun 4 '13 at 22:01