CS395T/CAM383M Computational Statistics > HW 5 John Woods HW5
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#1
03-04-2010, 02:42 PM
 johnwoods Member Join Date: Jan 2010 Posts: 30
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()
As you can see, U is the sum of the indicator variables Ij (as defined in the homework assignment). Thus, $\Large E[U]=\sum_{j}{E[I_j]}$.

What is the expectation of that indicator variable? It's a boolean, so the expression is $\Large 1 P[A] + 0 P[A^c]=P[A]$. Thus we define $\Large E[I_j]=P[A_j]$.

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 $\Large \frac{N-1}{N}$, so the probability of Aj is that to the Nth power.

Thus, we get $\Large E[U] = \sum_{j}{P[A_j]} = N\left(\frac{N-1}{N}\right)^N$.

Since U+V=N, E[U+V]=E[N]=N, and from that, E[U]+E[V]=N. This leaves us with E[V]=N-E[U], which we can write out in its entirety as: $\Large E[V] = N\left(1-\left(1-\frac{1}{N}\right)^N\right)$

Compute var[V]
First of all, we know from before that $\Large E[I_j] =P[A_j] = \left(\frac{N-1}{N}\right)^N$. It should also be clear that $\Large P[A_j\cupA_i] = \left(\frac{N-2}{N}\right)^N$, unless i=j, in which case it's just P[Aj].

We note that $\Large var[V]=var[U]$.

Further, we know U, so we can figure out that $\Large U^2 = \sum_{j}{I_j^2} + \sum_{i\ne j}{I_i I_j}$.

From that, let's get an expectation, and note that $\Large E[{I_i}^2] = E[I_i]$.

$\Large E[U^2] = \sum_{i}{E[{I_i}^2]}+\sum_{i\ne j}{\left(1-\frac{2}{N}\right)^N} = N\left(1-\frac{1}{N}\right)^N + N\left(1-\frac{2}{N}\right)^N$

$\Large E[V]^2$ is trivial to calculate. We can now give an expression for the variance of V:

$\Large var[V] = N\left(\left(1-\frac{1}{N}\right)^N + \left(1-\frac{2}{N}\right)^N\right) - N^2\left(1-\left(1-\frac{1}{N}\right)^N\right)^2$

Bonus
This is the easy part. First of all, 1/N times E[V] is $\Large \left(1-\left(1-\frac{1}{N}\right)^N\right)$. As N approaches infinity, the $\Large 1-\frac{1}{N}$ 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; 03-04-2010 at 02:45 PM. Reason: Wanted to save before I added bonus answer.
#2
05-10-2010, 10:37 AM
 jhussmann TA Join Date: Jan 2009 Posts: 76

You should reconsider your reasoning for the bonus question.

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