Question

Please write in detail neatly. Kindly don't use any symbols and shortcut words. Please write the formula appropriately. The question is:

Consider an unbiased 4-sided die where the sides are numbered 1, 2, 3, 4, and a biased coin with probability of head P(H) =4 divided by 7. A chance experiment consists of rolling the die once and then tossing the coins many times as the number showing on the die. Let X represent the outcome of the roll of the die, and let Y represent the number of heads observed after tossing the coin. Below, find each of the following:

a. R(X) and R(Y)

b. Give the distribution of X. What is the name of this distribution?

c. What is the name of the conditional distribution P(Y | X = 2). Give this distribution.

d. Give the conditional expectation E[Y | X = 2].

e. Give the joint distribution of the random vector < X, Y >.

f. Give P(Y = 3).

g. Give P(X = 2 | Y = 2).

h. Give E[E[Y|X]]. ( Kindly break down this formula and how to calculate it. please write in detail).

Please explain to me about "unbiased 4-sided die" and "a biased coin." I just need to understand the meaning of them. Could you please give two examples so that I can understand easily?

Answer 1

Unbiased 4-sided die means all of the four outcomes are equally likely. Biased coin means probability of getting head is 4/7 instead of 0.5.

(a)

R(X) shows the random variable X. Since die is unbiased so X can take values 1, 2, 3, and 4 with equal probabilities 1/4 = 0.25.

R(Y) shows the random variable Y. Since coin is biased and Y can take value head (H) and tail (T) with probabilities P(H) = 4/7 = 0.5714 and P(T) = 1-0.5714 = 0.4286 respectively

(b)

X has discrete uniform distribution. The pdf of X is

(c)

When X=2, coin is tossed 2 times. That is Y can values 0, 1, and 2. The pdf of Y is

P(Y=2 | X=2) = P(H)P(H) = 0.5714 *0.5714 = 0.32649796

P(Y=1 | X=2) = P(H)P(T) +P(T)P(H) = 0.5714 *0.4286 + 0.5714 *0.4286 = 0.48980408

P(Y=0 | X=2) = P(T)P(T) = 0.4286*0.4286= 0.18369796

d)

The expected value is

E(Y | X=2) = P(Y=2 | X=2) *2 + P(Y=1 | X=2) *1+P(Y=0 | X=2) *0 = 0.32649796 * 2 + 0.48980408* 1 + 0.18369796*0=1.1428

Answer: 1.1428

e)

When X=1, coin is tossed 1 time. That is Y can values 0, 1. The pdf of Y is

P(Y=1 | X=1) = P(H) = 0.5714

P(Y=0 | X=1) =P(T)= 0.4286

So,

P(Y=1 and X=1) = 0.5714* 0.25 = 0.14285

P(Y=0 and X=1) = 0.4286 * 0.25 = 0.10715

----------------------

When X=2, coin is tossed 2 times. That is Y can values 0, 1, and 2. The pdf of Y is

P(Y=2 | X=2) = P(H)P(H) = 0.5714 *0.5714 = 0.32649796

P(Y=1 | X=2) = P(H)P(T) +P(T)P(H) = 0.5714 *0.4286 + 0.5714 *0.4286 = 0.48980408

P(Y=0 | X=2) = P(T)P(T) = 0.4286*0.4286= 0.18369796

So,

P(Y=2 and X=2) = P(Y=2 | X=2)P(X=2) = 0.32649796 *0.25 = 0.08162449

P(Y=1 and X=2) = P(Y=1 | X=2)P(X=2) = 0.48980408*0.25 = 0.12245102

P(Y=0 and X=2) = P(Y=1 | X=2)P(X=2) = 0.18369796*0.25 = 0.04592449

Basically distribution of P(Y=X=k) , k=1,2,3,4 will be binomial distributions with parameter n=k and p=0.5714.

Following table shows the joint pdf:

		X
		1	2	3	4	P(Y=y)
	0	0.10715	0.04592449	0.019683236	0.008436235	0.181194
	1	0.14285	0.12245102	0.078723761	0.044988005	0.389013
Y	2	0	0.08162449	0.104952769	0.089965514	0.276543
	3	0	0	0.046640234	0.079960016	0.1266
	4	0	0	0	0.026650229	0.02665
	P(X=x)	0.25	0.25	0.25	0.25	1

f)

P(Y=3) = 0.1266