2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.