The problem goes as following:
Let there be $n$ coupons, and $X_i$ be the random variable whose value is $1$ if coupon $i$ is collected during the first $n$ draws, and $0$ otherwise. What is the expected value $E(X)$? and are $X_a$ and $X_b$ independent if $a\neq b$?
For the first part I got $0.632n$ but I'm not sure if it's correct. But I'm more or less stuck on the second part. It seems to me that they should be independent since each draw is independent of one another, but it also seems to me that drawing one coupon reduces the chance of drawing another one. Can someone help me with this problem?
Thanks everyone.
I assume that $X$ is the total number of distinct coupons drawn in the first $n$ rounds. Then $X=X_1+\cdots +X_n$, and therefore $E(X)=E(X_1)+\cdots+E(X_n)$.
The probability coupon $i$ is not drawn in the first $n$ rounds is $\left(\frac{n-1}{n}\right)^n$, so the probability it is drawn is $1-\left(\frac{n-1}{n}\right)^n$. This is $\Pr(X_i=1)$, which is equal to $E(X_i)$.
It follows that $E(X)=n\left(1-\left(\frac{n-1}{n}\right)^n\right)$.
Note that $\left(\frac{n-1}{n}\right)^n=\left(1-\frac{1}{n}\right)^n$, and for $n$ of even medium size this is quite close to $e^{-1}$. Thus unless $n$ is quite small, $E(X)\approx n\left(1-e^{-1}\right)$. This is very close to your $0.632n$.
As to independence, the intuition is clear: if coupon $a$ is drawn in the first $n$ rounds, the probability coupon $b$ is also drawn is less than the unconditional probability that coupon $b$ is drawn in the first $n$ rounds. So $X_a$ and $X_b$ cannot be independent.
We can if we wish confirm by calculating. The probability that neither $a$ nor $b$ is drawn is $\left(\frac{n-2}{n}\right)^n$. The product of the probability that $a$ is not drawn and the probability that $b$ is not drawn is $\left(\frac{n-1}{n}\right)^{2n}$, which is not equal to $\left(\frac{n-2}{n}\right)^n$. It follows that $\Pr(X_a=0\cap X_b=0)\ne \Pr(X_a=0)\Pr(X_b=0)$, and therefore $X_a$ and $X_b$ are not independent.