Question

Let XX be a random variable that is equal to the number of heads in two flips of a fair coin. What is \text E[X^2]E[X 2 ]

Answer 1

Answer:

Step-by-step explanation:

From the given information, it is likely that the random variable(X) have the values below:

Let head be H

Let tail be T

So;

X(HH) = 2;

X(HT) = 1;

X(TH) = 1;

X(TT) = 0

The distribution can now be computed as:

$p(X= TT) = \dfrac{1}{4}$

$p(X=TH) = \dfrac{1}{4}$

$p(X=HT) = \dfrac{1}{4}$

$p(X=HH)= \dfrac{1}{4}$

Now, the expected value that is equivalent to the number of heads when the coin is flipped twice is:

$E(X) = p(TT)*X(TT)+p(TH)*X(TH)+p(HT)*X(HT)+p(HH)*X(HH)$

$E(X) = \dfrac{1}{4}\times 0 + \dfrac{1}{4}\times 1 + \dfrac{1}{4}\times 1 + \dfrac{1}{4}\times 2$

$E(X) = 0 + \dfrac{1}{4}+ \dfrac{1}{4} + \dfrac{1}{2}$

$E(X) =\dfrac{1+1+2}{4}$

$E(X) =\dfrac{4}{4}$

E(X) = 1

$E(X^2) = p(TT)*X(TT)^2+p(TH)*X(TH)^2+p(HT)*X(HT)^2+p(HH)*X(HH)^2$

$E(X^2) = \dfrac{1}{4}\times 0^2+ \dfrac{1}{4}\times 1^2 + \dfrac{1}{4}\times 1^2 + \dfrac{1}{4}\times 2^2$

$E(X^2) = 0 + \dfrac{1}{4}+ \dfrac{1}{4} + \dfrac{4}{4}$

$E(X^2) =\dfrac{1+1+4}{4}$

$E(X^2) =\dfrac{6}{4}$

$E(X^2) =1.5$

Finally; To compute E²[X]

E²[X] = E[X]²

E²[X] = 1²

E²[X] = 1