coin toss and conditional expectation

Question

Very basic question, yet I'm confused since my results are different from what I would expect. Coin toss exercise: let $X_n = sum_{n}{B_i}$ is a random variable where $B_i=1$ if it is head and -1 if tail. Find $E[X_4|I_1]$. Possible values of $X_4$ are 4,2,0,-2,-4. Thus I write $E[X_4|I_1](+1)=4*P(X_4=4 \bigcap X_1=1)/P(X_1=1) + 2*P(X_4=2 \bigcap X_1=1)/P(X_1=1) - 2*P(X_4=-2 \bigcap X_1=1)/P(X_1=1) - 4*P(X_4=-4 \bigcap X_1=1)/P(X_1=1)=4*(0.5)^2 + 2*(0.5) - 2*(0.5)^2 - 0 = 3/2$ where there is 0.5 in the second term since there are two paths which lead to 2: HTH and THH. A similar expression for the case $X_1=-1$ where I get -3/2. Yet in the answers it is written $E[X_4|I_1]=X_1$. Could you tell me where is a mistake in my logic?

Answer 1

If $I_1$ is the $\sigma$-algebra generated by the first outcome, i.e. $I_1 = \sigma(X_1)=\sigma(B_1)$, then $$ E(X_4 \mid I_1) = E(B_1 + B_2 + B_3 + B_4 \mid 1_1) =E(B_2 + B_3 + B_4 \mid 1_1)+B_1 $$ Since $B_1$ is $I_1$-measurable (you know its value at trial $1$). Furthermore $$ E(B_2 + B_3 + B_4 \mid 1_1)+B_1 =E(B_2 + B_3 + B_4)+B_1 =B_1 = X_1 $$ Since the coin tosses are independent of the previous outcomes.