Edit: Dumb question from me. Still do not know exact formal way of solving but, since X gives you number of flips, of course number of heads is expected as half of it. However, what about variance?
I had a homework problem, which has an official solution released by teaching assistant. It's hard for me to reach him, so I ask you. Thank you!
Here is the question
- In a game, first, a die is rolled and then a coin is flipped the number of times given by the die. Compute the mean and variance of number of heads.
So, it is obvious that it is easy to solve it by conditioning and from there, find expected and variance. This makes sense for me.
However, the problem I have is that, TA did not explain everything and I am not good at probability unfortunately. He directly gives this information below;
- Given X = rolled a die, Y = number of heads. We know that $$E[Y|X=x] = \frac{x}{2}$$ and $$Var[Y|X=x]= x*\frac{(1^2 * 1/2 - (1 * 1/2)^2)}) = \frac{x}{4}$$
How can I derive this mean and variance? I tried by conditioning formulas but it is not easy since joint distribution or even marginal X distribution is not clear to me... Any help or idea is appreciated!
I tried to calculate die distribution by binomial but it is confusing...
Note: Something gone wrong with variance in terms of Mathjax. Sorry about that!
Let's think about the possible cases from rolling a die:
$ X = 1, 2, 3, 4, 5,$ or $6$
$Y = \frac{X}2$ thus $ Y = 0.5, 1, 1.5, 2, 2.5,$ or $3$ (all with equal probability since all faces of a fair die pop up with equal probability)
$E[Y] = \frac16(0.5+1+1.5+2+2.5+3) = 1.75$
$Var(Y) = \frac16((0.5-1.75)^2+(1-1.75)^2+(1.5-1.75)^2+(1.5-1.75)^2+(2.5-1.75)^2+(3-1.75)^2) = 0.729$