A series of Bernoulli trials with successful rate p (0; 1) is performed. We will stop the experiment whenever a changeover occurs, which means that the outcome differs from the one preceding it. Let X be number of Bernoulli trials being performed. (1). Prove that P{X >= 3} >= 0.5 (2). Find E[X].
I have proved the 1st by finding P{X < 3} = 2p^2-2p + 1 >= 0.5
How do I find E[X]? Thank you
Let $E_0$ denote the expected value given that the first trial resulted in $0$.
Let $E_1$ denote the expected value given that the first trial resulted in $1$.
Starting with the first trial we see that $$E=(1-p)E_0+pE_1$$
To compute $E_0$: considering the second trial we see that $$E_0=p\times 1+(1-p)(E_0+1)\implies E_0=\frac 1p$$
To compute $E_1$: considering the second trial we see that $$E_1=(1-p)\times 1+p(E_1+1)\implies E_1=\frac 1{1-p}$$
Thus $$E=\frac {1-p}p+\frac p{1-p}$$
Note: I don't follow your argument that $P(X≥3)≥.5$. You claim to have shown that $P(X<3)≥.5$ but this would imply the opposite of what you want. But to show the desired inequality just remark that $X≥3$ just means that the first two trials have to give the same answer so $P(X≥3)=p^2+(1-p)^2≥ .5$ Of course, that polynomial is the same as the one you wrote down, but you claim that it equals $P(X<3)$.