Binomial Random Variable and Bernoulli trials problem

Question

Let X be a Binomial random variable defined as the sum of 6 independent Bernoulli trials. The probability of a Bernoulli taking the value 1 is given by p. Suppose that prior to the 6 Bernoulli trials, p is chosen to take one of three possible values with the following probabilities:

p      Probability
---    -----------
0.2    0.1
0.6    0.7
0.8    0.2

• Compute the joint probability distribution of X and p. Are X and p independent?

• Compute the unconditional mean and variance of X.

• Compute the conditional mean of X given each possible value of p. Based on your calculations, what sign do you expect the covariance between X and p to be?

I'm having trouble understanding the lead-up of the question. I'm interpreting it that the probability of X is dependent on p, which also has probabilities and this is confusing me. What would the sample space of X be and how could you find the probabilities of X being values other than 1? Also for the probability of X being 1, how do you know which p value to use (0.2, 0.6, 0.8 from the table)? Thanking you in advance!

Answer 1

$X$ is the count of successes among $6$ Bernoulli trials, with success parameter $p$, while $p$ is itself a random variable with a support of three discrete values. $$P(p=0.2)=0.1, P(p=0.6)=0.7, P(p=0.8)=0.1$$

Thus the conditional distribution of $X$ for a given $p$ is binomial.

$$X\mid p=y \;\sim\; \mathcal{Binomial}(6, y)\\ \mathsf P(X=x\mid p=y) = \binom{6}{x}y^x(1-y)^{6-x}$$

So then $X$ itself has probability mass function:

$$\begin{align} \mathsf P(X=x) = {} & \mathsf P(X=x\mid p=0.2)\mathsf P(p=0.2) \\ & {} +\mathsf P(X=x\mid p=0.6)\mathsf P(p=0.6) \\ & {} +\mathsf P(X=x\mid p=0.8)\mathsf P(p=0.8) \end{align}$$

Substitute and simplify this, and determine if $\mathsf P(X=x, p=y) = \mathsf P(X=x)\mathsf P(p=y)$ for all $y\in \{0.2, 0.6, 0.8\}$

Answer 2

The joint distribution of the random variables $X$ and $P$ is given by $$pr(X=x,P=p) = pr(X=x\ |\ P=p) \ pr(P=p),$$ where $$pr(X=x\ |\ P=p) = \binom{6}{x}p^x (1-p)^{6-x}\ (x=0,1,2,3,4,5,6) $$ and $pr(P=p)$ is given by your table, for $p=0.2, 0.6, 0.8$.

The unconditional mean and variance of $X$ is found from the marginal distribution $$pr(X=x) = \sum_{p} pr(X=x,P=p). $$

This should be enough to answer all of the stated questions.