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A total of n balls, numbered 1 through n, are put into n urns, also numbered 1 through n in such a way that ball i is equally likely to go into any of the urns 1, 2, . . . , i.

Find

(a) the expected number of urns that are empty; (b) the variance of the number of empty urns.

(a)Let $X_i$ be a random variable that is 1 if the i-th urn is empty, 0 otherwise.

$X=\sum_{i=1}^N X_i$=number of empty urns.

$E[X]=\sum_{i=1}^N E[X_i]=\sum_{k=1}^N \frac{i-1}{N}=\frac{N-1}{2}$

(b) $$\begin{align} Var (X) &=Var (\sum_{i=1}^NX_i)\\ &=\sum_{i=1}^N Var (X_i)+2*\sum_{}^ {}\sum_{iI have a doubt about the subscript of the cov

for $i

$$\begin{align}E[X_i X_j] &= P(X_i=1,X_j=1)\\ &=P(X_i)P(X_j=1|X_i=1)\\ &=\prod_{i=1}^{j-1} (1-\frac{1}{i})*\prod_{j=1}^{N} (1-\frac{2}{j})\end{align}$$

$$\begin{align}Cov(X_i,X_j) &= E[X_i X_j]-E[X_i]*E[ X_j]\\ &=\prod_{i=1}^{j-1} (1-\frac{1}{i})*\prod_{j=1}^{N} (1-\frac{2}{j})-\prod_{i=1}^{j-1} (1-\frac{1}{i})*\prod_{j=1}^{N} (1-\frac{1}{j})\end{align}$$

1 Answers 1

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We can write the number of urns that are empty as $$ N_E = \sum_{j=1}^nX_i $$ where $X_i$ is an indicator that urn $i$ is empty, so that taking expected value of both sides gives $$ E(N_E) = \sum_{i=1}^nP(\mbox{urn i is empty}).$$ The probability that urn $i$ is not assigned ball $k$ is $1-1/k$ if $i\le k$ or else $1.$ Since the balls are independently placed in the urns, the probability that all $n$ balls are not placed in urn $i$ is $$ P(\mbox{urn i is empty})=\prod_{k = i}^n\left(1-\frac{1}{k}\right) =\prod_{k = i}^n\frac{k-1}{k} = \frac{i-1}{n}$$ where we used the fact that the product telescopes.

So for the expected value, we have $$E(N_E) = \sum_{i=1}^n\frac{i-1}{n} = \frac{n-1}{2}.$$

For the variance, we are focused on $$ N_E^2 = \sum_{i=1}^n\sum_{j=1}^n X_iX_j= \sum_{i=1}^nX_i^2 + 2\sum_{i

The only thing we haven't computed yet is $P(\mbox{urns i and j are empty})$ for $i

We can then compute $$ \sum_{i

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    Sorry but could you check what I have written, the solution is clear to me, but I'm not sure about the subscripts – 2017-01-25
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    @Anne In your problem setup you say there are $i$ urns, but you assume there are $N$ urns (so I changed it in mine). Also you never use the fact that there are $n$ balls. Your calculation of the expected value is impossible to follow after $E(X)=\sum_iE(X_i)$. Why does the index change to $k$? Why is $E(X_i)=(i−1)/N$? (it's not). If that were correct, why would the sum $=(N-1)/2$? . Likewise your *setup* for the variance calculation is correct, up to the line $E(X_iX_j)=P(X_i=1,X_j=1)$ but everything else makes no sense (what's k?). Things should not change with $i$... the urns are symmetric – 2017-01-25
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    can you write them in right form? – 2017-01-25
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    @Anne Whoops, looking carefully back at the problem I realize I made a big error in interpretation (I thought that each urn was equally likely to get any ball, not that ball $i$ needed to go to urns 1,2,..i). That makes what I said about symmetry, and my solution wrong. Makes your work make more sense, though I'm still not sure what $k$ is. Will look at this later and correct my answer. – 2017-01-25
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    @Anne K fixed mine. The form of yours makes more sense now, but yes, the indices are off. As it stands in your covariance formula you have $i$ and $j$ as free indices but take the product over them on the RHS. I calculated $E(X_iX_j)$ for $i – 2017-01-26