expectations

[rohan]

Suppose you toss an unbiased coin infinitely many times.

Let Xhh be the no. of tosses it takes to see two consecutive heads for the first time.
Let Xht be the no. of tosses it takes to see a tail right after the head for the first time.

Find E(Xhh) and E(Xht).

[Sinistral]

For Xhh I am getting a Fibonacci series.

E[Xhh]= 2*(0.25) +  Σ(r=3 to ∞) r*[0.5^(r-2) * t(r-2) * 0.25]
where t(i) is the i th term of a Fibonacci series.
i wasnt able to sum it!!

for E(Xht) i was able to create a summable series.
E[Xht]= 0.25 * Σ(r=2 to ∞) [r(r-1)/2^(r-2)]
E[Xht]= 0.25 * Σ(r=2 to ∞) {r^2/[2^(r-2)] - r/[2^(r-2)]}
both these terms r^2/[2^(r-2)] & r/[2^(r-2)] can be summed individually.

[Amit Goyal]

Xhh is the no. of tosses it takes to see two consecutive heads for the first time.
E(Xhh) = Pr(1st toss is H)E(Xhh|1st toss is H) + Pr(1st toss is T)E(Xhh|1st toss is T)
It is clear that
E(Xhh|1st toss is T) = 1+E(Xhh)
E(Xhh|1st toss is H) = Pr(2nd toss is H)E(Xhh|1st toss is H, 2nd toss is H) + Pr(2nd toss is T)E(Xhh|1st toss is H, 2nd toss is T)
E(Xhh|1st toss is H, 2nd toss is H) = 2
E(Xhh|1st toss is H, 2nd toss is T) = 2+E(Xhh)
Thus,
E(Xhh|1st toss is H) = 0.5(2) + 0.5(2+E(Xhh)) = 2 + 0.5E(Xhh)
E(Xhh|1st toss is T) = 1+E(Xhh)
E(Xhh) = 0.5(2+0.5E(Xhh)) + 0.5(1+E(Xhh))
E(Xhh) = 6

Similarly,
Let Xht be the no. of tosses it takes to see a tail right after the head for the first time.
E(Xht) = Pr(1st toss is H)E(Xht|1st toss is H) + Pr(1st toss is T)E(Xht|1st toss is T)
It is clear that
E(Xht|1st toss is T) = 1+E(Xht)
E(Xht|1st toss is H) = Pr(2nd toss is H)E(Xht|1st toss is H, 2nd toss is H) + Pr(2nd toss is T)E(Xht|1st toss is H, 2nd toss is T)
E(Xht|1st toss is H, 2nd toss is H) = 1+E(Xht|1st toss is H)
E(Xht|1st toss is H, 2nd toss is T) = 2
Thus,
E(Xht|1st toss is H) = 0.5(1+E(Xht|1st toss is H)) + 0.5(2)
gives us
E(Xht|1st toss is H) = 3
E(Xht|1st toss is T) = 1+E(Xht)
E(Xht) = 0.5(3) + 0.5(1+E(Xht))
E(Xht) = 4