# Thread: Properties of discrete random variables

1. ## Properties of discrete random variables

Hello,

would some one with more experience mind to confirm or correct me:

For discrete random variables I know:

E[aX+b] = aE[X]+b and Var(aX+b)=a^2Var(X)+b

if there are more than two constants, is correct to assume that this holds?

E[abX+c] = abE[X]+c and Var(abX+c) = (ab)^2Var(x)+b

Many Thanks
jc

2. ## Re: Properties of discrete random variables

The variance of a constant is 0.

V(aX+b) = a^2 V(X)

V(abX + c) = (ab)^2 V(X)

3. ## Re: Properties of discrete random variables

Thanks for that

4. ## Re: Properties of discrete random variables

It's easy to see that
V(abX + c) = (ab)^2 V(X) from
V(dX + c) = (d)^2 V(X)
just by letting ab = d. Since a and b are constants then d = ab is a constant so we get the result for the two constants as a direct consequence.

5. ## Re: Properties of discrete random variables

Hi,

I have to catch up on that, there is still some confusion.

I have a random varialbe X with E[X]=7.45 min. On a daily basis I am charged 0.3 per min and fix costs of 3.

So for a day E[X] = 0.3*7.45 + 3 = 5.235

But for a year (300 days)?

My intution would tell me 300*(0.3*7.45+3)=1570.5,

but following the formula E[abX+c] = abE[X]+c strictly, I would get :

0.3*300*7.45+3 = 673.5, which seems to low to me.

I am going off in a totally wrong direction?