The variance of a constant is 0.
V(aX+b) = a^2 V(X)
V(abX + c) = (ab)^2 V(X)
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
The variance of a constant is 0.
V(aX+b) = a^2 V(X)
V(abX + c) = (ab)^2 V(X)
Thanks for that
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.
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?
Thanks a lot for comments.
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