The question as follows: There's a gambling game with 100 rounds. First, a random number is randomized out of unknown continuous distribution. then, each round a random variable is randomized from same distribution. There are 2 gamblers: gambler A gambles the next random number would be bigger than the previous.
Question: what is the Expectation of number of rounds when gambler A was right?
The answer: $suppose\ X_{i}\ is\ the\ value\ of\ the\ number\ randomized\ in\ round\ i$ $and\ Y_{i}\ is\ an\ indicator\ that\ X_{i}>X_{i-1}:$
The Equation is as Follows:
$E(Y)=E(\sum_{i=1}^{100}Y_{i})=\sum_{i=1}^{100}E(Y_{i})=\sum_{i=1}^{100}P(X_{i}>X_{i-1})=\sum_{i=1}^{100}\frac{1}{2}=50$
and it's said that $P(X_{i}>X_{i-1})=\frac{1}{2}$ due to symmetry. I can't get it.
we are expected to show in our answers the whole way we get to the solution, with explanation for every equation.
I tried to convert P in terms of cumulative $F_{X}(x)$ but I'm confused which is my integration parameter and so on.
Thanks to everybody who tries to help!
The simple answer is due to symmetry and independence of the random variables.
The Formulaic derivations:
$P(X_{i}>X_{i-1})\underset{(1)}{=}\int_{-\infty}^{\infty}P(X_{i}>x|X_{i-1}=x)\cdot f_{X_{i-1}}(x)dx\underset{(2)}{=}\int_{-\infty}^{\infty}(1-F_{X_{i}}(x))\cdot f_{X_{i-1}}(x)dx\underset{(3)}{=}\int_{-\infty}^{\infty}(1-F_{X_{i-1}}(x))\cdot f_{X_{i}}(x)dx\underset{(4)}{=}P(X_{i-1}>X_{i})$
1.Law of total probability
2.independence+ tail of comulative function
3.same dist.
4.obvious