Answer to Question #242099 in Statistics and Probability for HRr

Question

Let X be a Bernoulli random variable (Hint: Special case of Binomial

Distribution)

a. Compute E(X²)

b. Show that V(X) is p(1 − p).

c. Compute E(X⁷⁹)

Answer 1

If X is A Bernoulli random variable, then its p. d. f is given by

$P(X=x)=p^{x}(1-p)^{1-x}, x=0, 1$

First, determine P(X=x) when x=0 and x=1.

So,

P(X=0)=p⁰(1-p)^1-0=1*(1-p)=1-p

P(X=1)=p¹(1-p)^1-1=p*1=p

a.

E(X²)=$\displaystyle\sum_{x=0}^{1}(\def\foo{x^2}\foo)*P(X=x)$

=0²*P(X=0)+1²*P(X=1)

=0*(1-p)+1*(p)=p

Therefore, E(x²)=p

b. Variance of random variable X is defined by,

V(X)=E(X²)-(E(X))²

We need to determine E(X) which is given as,

E(X)=$\displaystyle\sum_{x=0}^{1}(x)*P(X=x)$

=0*P(X=0)+1*P(X=1)

=0*(1-p)+1*(p)=p

Therefore, V(X)=p-p²

Factoring gives p(1-p)

Hence V(X)=p(1-p) as given.

c.

E(X⁷⁹) =$\displaystyle\sum_{x=0}^{1}(x^{79})*P(X=x)$

=(0⁷⁹)*P(X=0)+(1⁷⁹)*P(X=1)

=0*(1-p)+1*p

=p