# Can someone just explain this part of the solution to me....

• Apr 10th 2009, 11:50 AM
zhupolongjoe
Can someone just explain this part of the solution to me....
It says

Suppose P(X=a)=p and P(X=b)=1-p
So clearly (X-b)/(a-b) is a Bernoulli Random variable. Find Var(X).

The solution says:

p(1-p)=Var((X-b)/(a-b))=1/(a-b)^2Var(X-b)=1/(a-b)^2*Var(X)

So

Var(X)=(a-b)^2*p(1-p)

What I don't understand is how they did the part where I have underlined the equal sign (i.e. how to get out the 1/(a-b)^2 from that). Thanks.
• Apr 10th 2009, 11:54 AM
Moo
Hello,
Quote:

Originally Posted by zhupolongjoe
It says

Suppose P(X=a)=p and P(X=b)=1-p
So clearly (X-b)/(a-b) is a Bernoulli Random variable. Find Var(X).

The solution says:

p(1-p)=Var((X-b)/(a-b))=1/(a-b)^2Var(X-b)=1/(a-b)^2*Var(X)

So

Var(X)=(a-b)^2*p(1-p)

What I don't understand is how they did the part where I have underlined the equal sign (i.e. how to get out the 1/(a-b)^2 from that). Thanks.

For any constant a, $\displaystyle \text{Var}(aX)=a^2 \text{Var}(X)$

We indeed have :
$\displaystyle \text{Var}(X)=E(X^2)-[E(X)]^2$

So $\displaystyle \text{Var}(aX)=E(a^2X^2)-[E(aX)]^2$

But for any constant b, $\displaystyle E(bX)=bE(X)$

So $\displaystyle \text{Var}(aX)=a^2 E(X^2)-a^2 [E(X)]^2=a^2 [E(X^2)-[E(X)]^2]=a^2 \text{Var}(X)$
• Apr 10th 2009, 12:38 PM
zhupolongjoe
Ah thanks. I don't recall that formula from class so I was confused.