Expected Value and Variance - May 24

[Amit Goyal]

You roll a fair six-sided die, and then you flip a fair coin the number of times shown by the die. Find the expected value and the variance of the number of heads obtained.

[aditi5000]

E (H given the no. of flips) = np= 1/2 * no of flips = 1/2 * (1+6)/2 = 7/4
variance = npq= no. of flips * 1/2 * 1/2 = 1/4 *[ 1/12 * (36-1) +  7/2] = 77/ 48

[Badmathsboy]

Aditi, please help. Still not getting it.

[AJ]

Hello Sir,

I am getting probability of H=r as this:


by listing down all the options.
But to find expectation of H, of-course this becomes really messy. Please give a hint about how to proceed.

[Amit Goyal]

You can use indicator functions (the way we did when we found expected number of true propositions). Like: Let I_j takes value 1 if the jth toss is head and 0 otherwise. And then N, number of heads  = I_1 + I_2 + I_3 + I_4 +I_5 + I_6
Then use linearity of expectation. And rest is just simple computing.

[aditi5000]

Hey bad maths boy, I used the fact that such an experiment would be a uniform distribution.. (discrete)

[AJ]

@Amit sir,

Is, Expectation= 7/4  ...?
Variance= zero

I am confused in variance..
Are these correct, If not please tell me, I will try this again...

[AJ]

NO,
Variance= 161/144

[Amit Goyal]

There is another computationally less involved way to do this problem:
Let X be the number of heads. Let N be the number on the dice.
E[X] = E[E[X|N]] = E[Np] = pE[N] = (1/2) (21/6) = 7/4
V[X] = E[V[X|N]] + V[E[X|N]] = E[Np(1-p)] + V[Np] = p(1-p)E[N] + (p^2)V[N] = 21/24 + 105/144 = 231/144 = 77/48