I have an Urn with 100 balls, 60 of them black. I start by taking 10 balls at random, taking note of how many black balls are in those 10 (let's call that $X_1$) and putting them back in the urn. Now I repeat the process taking out $X_1$ balls, noting the amount of black balls in those (call that $X_2$) and putting them back.
Now I consider the random variables $X_{n}$ defined as $X_{1} = 10 $ and for $n \geq 2$ $X_n =$ the amount black balls in the $(n-1)$th extraction.
Find $E(X_n)$ and $X_n \rightarrow _{a.s.} 0$
I was having some trouble with the first part, specifically if conditioning only to the last step is right.
My attempted solution
$$ E (X_n) = E( E(X_n | X_{n-1})) $$
Since $ X_{n} | X_{n-1} = k \sim H(100, 60, k) $ we have that $E(X_n | X_{n-1} = k ) = \frac{3}{5} k \{ X_{n-1} = k \} $. Then,
$$ E( E(X_n | X_{n-1})) = E(\frac{3}{5} X_{n-1}) = \frac{3}{5} E(X_{n-1}) $$
We know that $ E(X_{1}) = 6$, so we can compute $E(X_2) = 2 \frac{ 3^2}{5} $ and by induction we prove that for $ n \geq 2$, $E(X_n) = 2 \frac{3^n}{5^{n-1}} = 10(\frac{3}{5})^n$.
We want to prove that $P (Lim X_{n} = 0) = 1 $ then since $X_{n} \geq 0$ we need $Lim P ( X_{n} \geq \frac{1}{n} ) = 0 $. Using Markov's inequality we have that
$$ P( X_{n} \geq \frac{1}{n} ) \leq n E(X_{n}) = 10n (\frac{3}{5})^n \rightarrow 0 $$
So $P (Lim X_n = 0 ) = 1$.
Any comments would be greatly appreciated.
Your arguments are sound to me.
Another way to show that $X_n\to 0$ almost surely is the following. Let $Y_n$ be the number of blacks you get drawing always 10 balls, that is, after drawing $X_{n-1}$ balls, you draw another $10-X_{n-1}$. Clearly the event $\{X_n\geq k\}$ is included in the event $\{Y_n\geq k\}$, so $P(X_n\geq 1)\leq P(Y_n\geq 1)$, and if you intersect these events you maintain the inequality, that is
$$ P(X_1\geq 1\cap X_2\geq 1\cap\cdots \cap X_n\geq 1)\leq P(Y_1\geq 1\cap Y_2\geq 1\cap\cdots \cap Y_n\geq 1) $$ But the $Y_n$ are independent, so
$$ P(X_1\geq 1\cap X_2\geq 1\cap\cdots \cap X_n\geq 1)\leq P(Y_1\geq 1)^n=(1-P(Y_1=0)^n=(1-0.4^{10})^n $$ and clearly the last term approaches 0 as $n\to \infty$. Since $X_n$ is discrete and non increasing, $P(X_n\to 0)=P(\exists N: X_N=0)=1-P(X_n\geq 1,\forall n)$. Now, $\{X_n\geq 1, \forall n\}=\cap \{X_i\geq 1,i=1,\ldots,n\}$, which are nested events, so $P(X_n\geq 1, \forall n)=\lim_{n\to \infty} P(X_i\geq 1,i=1,\ldots,n)$, which we showed to be 0.