I have $f_{\tau, \theta}(y)=\theta e^{-\theta(y-\tau)}, y\ge\tau, \theta\gt 0$. Assume both parameters unknown.

$\mu_1=E(Y)=\tau+\frac1\theta=\bar{Y}=m_1$ where $m$ is the sample moment.

$\mu_2=E(Y^2)=(E(Y))^2+Var(Y)=(\tau+\frac1\theta)^2+\frac{1}{\theta^2}=\frac1n \sum Y_i^2=m_2$.

$\mu_2-\mu_1^2=Var(Y)=\frac{1}{\theta^2}=(\frac1n \sum Y_i^2)-{\bar{Y}}^2=\frac1n\sum(Y_i-\bar{Y})^2\implies \hat{\theta}=\sqrt{\frac{n}{\sum(Y_i-\bar{Y})^2}}$

Then substitute this result into $\mu_1$, we have $\hat\tau=\bar Y-\sqrt{\frac{\sum(Y_i-\bar{Y})^2}{n}}$

Is my procedure correct?


1 Answer 1


There is a small problem in your notation, as $\mu_1 =\overline Y$ does not hold. The term on the right-hand side is simply the estimator for $\mu_1$ (and similarily later). But your estimators are correct for $\tau, \theta$ are correct.

  • $\begingroup$ Maybe better wording would be "equating $\mu_1=m_1$ and $\mu_2=m_2$, we get ..."? $\endgroup$
    – CoolKid
    Oct 1, 2017 at 22:08
  • $\begingroup$ A better wording would be to first write $\theta = (m_2 - m_1^2)^{-1/2}$ and then write "plugging in the estimators for $m_1, m_2$ we get $\hat \theta = \ldots$". $\endgroup$
    – Dominik
    Oct 1, 2017 at 22:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .