#1




John Woods HW5
Compute the expected value of V.
I like thinking of this in terms of the code I wrote in Octave (Matlab). Here it is: Code:
N = 25 X = randsample(N,N,true) Yi = @(i) (sum(X == i)); Y = zeros(1,N); for i=1:N Y(i) = Yi(i); end V = @() ( sum(Y > 0) ); U = @() ( sum(Y == 0) ); Yi(5) V() U() What is the expectation of that indicator variable? It's a boolean, so the expression is . Thus we define . What is the value for this probability of Aj? The probability of some value i not showing up in a set of 1...N is , so the probability of Aj is that to the Nth power. Thus, we get . Since U+V=N, E[U+V]=E[N]=N, and from that, E[U]+E[V]=N. This leaves us with E[V]=NE[U], which we can write out in its entirety as: Compute var[V] First of all, we know from before that . It should also be clear that , unless i=j, in which case it's just P[Aj]. We note that . Further, we know U, so we can figure out that . From that, let's get an expectation, and note that . is trivial to calculate. We can now give an expression for the variance of V: Bonus This is the easy part. First of all, 1/N times E[V] is . As N approaches infinity, the term approaches 1 (approach is made more rapid by the exponent). Thus, the limit of the entire expression will be 0. Last edited by johnwoods; 03042010 at 02:45 PM. Reason: Wanted to save before I added bonus answer. 
#2




You should reconsider your reasoning for the bonus question.

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