Hey guys I have been working on a probability and expected value/variance problem and the problem is:
Each day the price of a stock in the market is a random number between 0 and 1 independently of its value in previous days. A record breaking price occurs when the price on the current day exceeds all previous prices. What is the expected number of record breaking prices in n days? What is the variance of the number of record breaking prices in n days? (Hint: If I throw k points at random in [0,1] what is the chance that the ith point I throw is the largest among those k points? Given that the ith point is the largest among those k points, is its distribution different from uniform in [0, 1]?)
My approach was: first day Expected v = 1*1*(1 choose 1) = 1 second day expected v = first day Expected V + P(second price>first price) * 2(2 choose 1) but then from third day, it becomes way too complicated.
I know I should be using the integral to find Expected value of continuous variable, but then I would need to know the probability of each cases, which I have no idea of figuring it out.
Intuitively I know that the probability of getting the max value gets significantly lower as days go on, and I would assume it's on the line of a log function. But again, this is just a guess.
Can someone please lend a hand? thank you!
I'm not sure about the variance, but here's an approach for the mean.
For $i=1,2,...,n$, let $X_i=1$ if there is a record set on day $i$, and $X_i=0$ if there isn't a record. For $X_i=1$, we must have that the value on day $i$ is the highest of the first $i$ days. So $P(X_i=1)=\frac1i$. So $E[X_i]=\frac1i$.
Since expectation is linear we have $E[\sum_{i=1}^n X_i]=\sum_{i=1}^n E[X_i]=1+\frac12+\cdots+\frac1n$, and this gives the expected number of records over $n$ days.
This confirms your idea that the number of records is roughly logarithmic in $n$.