Question

Suppose that

$P\{X=a\}=p,P\{X=b\}=1-p$

(a) show that $\frac{X-b}{a-b}$is a Bernoulli random variable

(b) Find Var(X).

Short answer

In the given information the answer os part (a) is$\frac{X-a}{b-a}~\left(\begin{array}{cc}0& 1\\ p& 1-p\end{array}\right)$which show that is Bernoulli random variable.

(b) is$Var\left(X\right)=(b-a{)}^{2}p(1-p)$

Step-by-step solution

 Step 1 :Given Information (Part-a)

With the probability $p$,random variable $X$assumes $\alpha$.Hence, with the same probability, random variable $\frac{X-a}{b-a}$assumes$0$

with the probability localid="1646975947088" $1-p$,random variable $X$assumes $b$Hence, with the same probability ,random variable $\frac{X-a}{b-a}$assumes$1$

:Explanation (Part-a)

$\frac{X-a}{b-a}~\left(\begin{array}{cc}0& 1\\ p& 1-p\end{array}\right)$

which shows that it is a Bernoulli random variable

:Final Answer (Part-a)

The answer is $\frac{X-a}{b-a}~\left(\begin{array}{cc}0& 1\\ p& 1-p\end{array}\right)$

which shows that$\frac{X-a}{b-a}$is Bernoulli random variable

:Given Information(Part-b)

We know that$Var\left(\frac{X-a}{b-a}\right)=p(1-p)$.

Step 5:Explanation (Part-b)

$Var\left(\frac{X-a}{b-a}\right)=\frac{1}{(b-a{)}^2}Var(X-a)=\frac{1}{(b-a{)}^2}Var\left(X\right)$

so we get that$Var\left(X\right)=(b-a{)}^{2}p(1-p)$

$Var\left(X\right)=(b-a{)}^{2}p(1-p)$

$Var\left(X\right)=(b-a{)}^{2}p(1-p)$

:Final Answer (Part-b)

$Var\left(X\right)=(b-a{)}^{2}p(1-p)$

The answer is$Var\left(X\right)=(b-a{)}^{2}p(1-p)$$Var\left(X\right)=(b-a{)}^{2}p(1-p)$

$Var\left(X\right)=(b-a{)}^{2}p(1-p)$$Var\left(X\right)=(b-a{)}^{2}p(1-p)$