A drunk person is in place $x=0$ in time $t=0$. Every second he moves forward one meter in a probability of $\frac{1}{2}$ or stays in his place in a probability of $\frac{1}{2}$ (his decisions are independent). In addition, every second he can pass out in a probability of $\frac{1}{2}$ (also independently).
Mark: $X$ - the distance he reaches.
A. Find the expected value of $X$.
B. Given he reaches at least 2 meters, find the expected value of $X$.
My solution:
A. $Y$ - number of seconds until he passes out. $Y\sim Geo(\frac{1}{2})$
$X|Y\sim Bin(y,\frac{1}{2})$
Therefore: $E(X) =E(E(X|Y))=E(\frac{Y}{2})=\frac{1}{2}\cdot E(Y)=\frac{1}{2}\cdot2=1$
B. $E(X|X\ge2)=\sum_{x=0}^{\infty}x\cdot P(X|X\ge 2)$
$P(X|X\ge 2)=\frac{P(X,X\ge2)}{P(X\ge2)}$
But I'm stuck at this point. Also, is it just $E(x)+2$?
Let E be the expected value of the distance after it has moved by 2 meters. Starting from 2, • We have a $\frac{1}{4}$ chance of passing out and staying put; from here the distance is $2$ . • We have a $\frac{1}{4}$ chance of not passing out and staying put and the distance then is E. • We have a $\frac{1}{4}$ chance of moving forward and not passing out and the distance is $(E+1)$. • We have a $\frac{1}{4}$ chance of moving forward and passing out and the distance is $3$. Thus, $$E = \frac{1}{4}(2) + \frac{1}{4}E + \frac{1}{4}(1 + E) +\frac{1}{4}(3)$$ Solving gives $E = 3$.