# Thread: expected value and variance (continuous random variables)

1. ## expected value and variance (continuous random variables)

Hello,

X1, X2, … are independent continuous random variables with E[x]=2, V(X)=9.

Yi=(0.5^i)Xi, i=1,2,…
Tn=Y1+Y2+…+Yn
An=(1/n) Tn

E and V of Yn, Tn, An ???

E[Yn] = E[(0.5^i)Xi] = (0.5^i)E[Xi] = 2(0.5^i)
V[Yn] = V[(0.5^i)Xi] = [(0.5^i)^2] V[Xi] = 9 [(0.5^i)^2]

2. YOU mean Yi not Yn
Then sum to get Tn, which is a partial geometric sum from calc 2.

3. Thanks, matheagle!

Is the following solution correct ?

E(Tn) = E(Y1)+E(Y2)+…+E(Yn) = 2*SUM(0.5^n)

V(Tn) = V(Y1)+V(Y2)+…+V(Yn)= 9*SUM((0.5^n)^2)

E(An)=E((1/n)* Tn)= 1/n *E(Tn) = 2/n* SUM(0.5^n)

V(An)= V((1/n)* Tn) =(1/n)^2 *V(Tn) = ((1/n)^2 )*(9*SUM((0.5^n)^2))

4. $T_n=\sum_{i=1}^n\left(1/2\right)^iX_i$

$E(T_n) =\sum_{i=1}^n\left(1/2\right)^iE(X_i) =2\sum_{i=1}^n\left(1/2\right)^i$

$V(T_n) =\sum_{i=1}^n\left(1/4\right)^iV(X_i) =9\sum_{i=1}^n\left(1/4\right)^i$

5. thanks for your response, matheagle!

The same solution I was thinking, but I could not insert the summation symbol, so I used the word "SUM" (from 0 to n) !!! Maybe I confused you, I apologize.

Anyway, thanks again.

6. yea, but not only did I use TeX, which you should learn
But I switched the powers so you can obtain the geometric sum.

7. I know it was my fault!