Number of random values that are recorded as being greater than all previous values Basically, given a continuous distribution with a density function f, record all numbers that are larger than all previous values.  (if the sequence 3, 2, 4, 3.5, 5 is generated, 3,4, and 5 are the recorded values).
I need to find the expected number of recorded values when 100 such values are generated.
My original thinking was that given a continious distribution from -infinity to infinity, there is a 50% chance that the next value will be larger than the current largest value, meaning that given the first value generated is automatically recorded, the remaining 99 values have a 50% chance to be larger, which would imply an estimated recorded value of 49.5 (rounded to 50) plus the first value for a grand total of 51 values.
After briefly discussing with my professor, this is apparently an incorrect way to go about the problem, and I am curious how I should go about solving this instead?  I assume there is some sort of diminishing returns on the chance that the number is largest, but as it is a continuous set and not a finite set, I don't follow why that would be the case.
 A: Let $X_i=1$ if at the $i$-th trial we get something bigger than all the previous ones, and $X_i=0$ otherwise.
Then the number of "biggests so far" is $X_1+X_2+\cdots+X_{100}$.
Now use the linearity of expectation to calculate $E(X_1+X_2+\cdots+X_{100})$.
To calculate $E(X_i)$, note that this is $\Pr(X_i=1)$.
Since the distribution is continuous, with probability $1$ al results are different. What is the probability that the $i$-th result is bigger than all the previous ones? To put it another way: look at the numbers obtained on trials $1$ to $i$. What is the probability the biggest among these occurs at the end?
This is where your intuition about "diminishing returns" gets an explicit number attached to it. 
Added: Let $n=100$. If $W$ is the number of values that we write down, then 
$$W=E_1+X_2+\cdots+X_n,$$
and therefore
$$E(W)=E(X_1+X_2+\cdots+X_n)=E(X_1)+E(X_2)+\cdots+E(X_n).$$
Among the $i$ values obtained in the first $i$ trials, the largest is equally likely to be in any of the $i$ places, so the probability it is at $i$ is $\frac{1}{i}$.
Thus $E(X_i)=(1)\frac{1}{i}+(0)\frac{i-1}{i}=\frac{1}{i}$. It follows that
$$E(W)=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}.$$
This number is the $n$-th harmonic number $H_n$.
