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?

  • $\begingroup$ 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 :-) $\endgroup$
    – hejseb
    Aug 10, 2013 at 20:36
  • $\begingroup$ 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. $\endgroup$
    – Djamillah
    Aug 10, 2013 at 20:52
  • 1
    $\begingroup$ Djamillah -- when you do that with expectations, it works because expectation is a linear operator. Variance isn't $\endgroup$
    – Glen_b
    Aug 11, 2013 at 0:20

1 Answer 1


The variance of the sum of two variables must be calculated with a term accounting for the covariance of those two variables.

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

Note that the coefficients on the variables are also squared in the first two terms of that equation. If X and Y are fully independent then the third term is zero of course.


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