# Calculating Expectation $\mathbb{E}Y_n$ of a random Variable.

Given $$n$$ balls inside a box, $$k$$ black and $$n-k$$ white. We're picking all the balls out and laying them in order. Let $$X_1,\ldots,X_n$$ be random variables such that $$X_i = 1$$ is the $$i$$-th chosen ball is black and $$X_i = 0$$ if white. Define the random variable $$Y_n = 1 + \sum_{i = 1}^{n-1} \mathbb{1}_{\{X_i \neq X_{i+1}\}}$$ where $$\mathbb{1}$$ is the indicator function. What's the expectation $$\mathbb{E}Y_n$$ and variance $$\text{Var}(Y_n)$$?

I'm new to probability and have been trying this problem for a while, still can't solve it. The probability distrubition of $$X_i$$ here is not Bernoulli distribution, because the probability $$p$$ of picking up a black ball in the $$i$$-th draw depends on results of $$1,\ldots, (i-1)$$ draws for $$i > 1$$.

Let $$M := \{\overbrace{1,\ldots,1}^{k}, \overbrace{0,\ldots,0}^{n-k}\}$$ be a multiset. We can think of our sample space as $$\Omega := \{\omega = (\omega^1,\ldots,\omega^n) : \omega^i \in M \}$$ with cardinality $$\frac{n!}{k!(n-k)!}$$.

Since $$Y_n(\omega) = \Big(1 + \sum_{i = 1}^{n-1} \mathbb{1}_{\{X_i \neq X_{i+1}\}}\Big)\omega$$, the possible numerical values for each $$Y_n$$ is the set $$1,\ldots,n$$. So $$\mathbb{E}Y_n = \sum_{x = 1}^{n} |x| \cdot \mathbb{P}(Y_n = x)$$

Now I am stuck calculating $$\mathbb{P}(Y_n = x)$$, it gets messy with different cases to consider. If $$x = 1$$, it must be that all the draws from $$1$$ to $$n$$ fail (no black ball), but that still depends on $$k$$..and so on, with more cases. What am I not seeing?

Any help would be really great!

EDIT:

Using @LostStatistician18's hint, we first have $$\mathbb{E}(\mathbb{1}_{\{X_i \neq X_{i+1}\}}) = 0 \cdot \mathbb{P}(X_i = X_{i+1}) + 1 \cdot \mathbb{P}(X_i \neq X_{i+1}) \\ = \mathbb{P}(X_i = 1, X_{i+1} = 0) + \mathbb{P}(X_i = 0, X_{i+1} = 1) = 2 \ \frac{k (n-k)}{n (n-1)}$$

for $$i = 1,\dots,(n-1)$$. So now we get $$\mathbb{E}(Y_n) = 1 + \sum_{i = 1}^{n-1} \mathbb{E}(\mathbb{1}_{\{X_i \neq X_{i+1}\}}) = 1 + \frac{2 k (n-k)}{n}$$

• What is $N$ here? Do you mean $n$ or is $N$ some random variable? May 22, 2020 at 16:59
• I mean $n$, edited it now!
– 9Sp
May 22, 2020 at 17:03

Just a hint: Although you are proposing to calculate the expectation using the definition $$E(Y) = \sum_{y \in Y(\omega)} y P(Y=y)$$, you might also try using the linearity of the expectation in this case.
$$E(Y) = 1 + \sum_{i=1}^{n-1} E 1_{\{X_i \ne X_{i+1}\}}$$ The expectation of the indicator variables can be calculated more easily as they only take the values 0 and 1. To calculate the variance, note that generally if $$Y= \sum_{i=1}^n X_i$$, then $$Var(Y) = \sum_{i=1}^n Var(X_i) + 2 \sum_{1\le i < j \le n} Cov(X_i,X_j).$$
In your case the $$X_i$$ are random variables taking the values 0 and 1, and so these variances/covariances can also be more easily calculated.
Going to the trouble of determining the entire distribution of $$Y$$ would be much tougher in this case (although could certainly be done too with some hard work).