Student
Hello Sir/Madam,
I just want to know please how to get the variance in this exemple :
Pθ = Exponential(θ), θ > 0 ; pdf(x) = 1 θ exp (−x/θ) if x ≥ 0, pdf(x) = 0 otherwise ; θb = X.
Thank you!
Teacher
Hello,
your question is quite hard to read. From what I understand, you ask how to get the variance of the mean \overline{X} (I imagine \theta b is the mean ?) in an exponential model of parameter \theta (please next time try to explain with words you question or if you can't, use latex your question is very hard to read).
Anyway, the variance of independent random variables is the sum of the variances of each random variable:
Var(\overline{X})=Var\left(\frac{1}{n}\sum_{i=1}^n X_i\right)=\frac{1}{n^2}Var\left(\sum_{i=1}^n X_i\right)=\frac{1}{n^2}\sum_{i=1}^n Var\left(X_i\right)
Be careful that this is true only because the X_i are independent.
Then, use that the variance of an exponential of parameter \theta is \theta^2 to conclude that Var(\overline{X})=\theta^2/n. (For the computation of the variance of an exponential, see the formula in slides 1, it is just some basic computation of integrals).
Best regards,
Timothée Mathieu