Calculate Variance for Exponential Distribution

I have a problem with calculating the variance of an exponential distribution. In my formulary there are these formulas for exponential distributions:

$E(X)=\frac{1}{\lambda}$ $V(X)=\frac{1}{\lambda^2}$

Where $E(X)$ is the expected value and $V(X)$ the variance.

I have a distribution function on the form $F_X(x)=C_1(1-e^{-\lambda_1x})+C_2(1-e^{-\lambda_2x})$

where the $C$s and $\lambda$s are constants.

I calculated $E(X)$ in the following way: $E(X)=C_1*\frac{1}{\lambda_1}+C_2*\frac{1}{\lambda_2}$

But when I try to calculate the variance using the formula above for V(X) and the same method as I used when calculating $E(X)$ I don't get the right answer.

I get the right answer if I use $V(X)=E(X^2)-(E(X))^2$,

but both should generate the same answer since $E(X^2)=\frac{2}{\lambda^2}$ (from partial integration) which gives $V(X)=\frac{2}{\lambda^2}-(\frac{1}{\lambda})^2=\frac{1}{\lambda^2}$

Can someone explain this please?

• Would you mind showing what you did when "using the formula above for $V(X)$"? It is usually easier to see what's gone wrong if you get to see what goes wrong :-) – hejseb Aug 10 '13 at 20:36
• I did like this: $V(X)=C_1*\frac{1}{\lambda^2}+C_2*\frac{1}{\lambda^2}$ I tried to square the C:s as well but that didn't work either. – Djamillah Aug 10 '13 at 20:52
• Djamillah -- when you do that with expectations, it works because expectation is a linear operator. Variance isn't – Glen_b Aug 11 '13 at 0:20

$$Var(aX + bY) = a^2Var(X) + b^2Var(Y) + 2ab Cov(X,Y)$$