In a textbook exercise, for a sequence of iid variables, I have calculated the score function to be $$\begin{bmatrix} - \frac{n}{2\lambda} + \sum_{i=1}^n \frac{( x_i - \mu)^2}{2\mu^2 x_i}& \sum_{i=1}^n \frac{\mu - x_i}{\mu^3} \end{bmatrix}$$ From this, the information matrix is readily calculated. I wish to show that the MLE estimators are consistent and asymptotically normal.

It is easy to see that the MLE estimator for $\mu$ is $\sum x_i/n$, and therefore the MLE estimator for $\lambda$ is $$ \frac{n}{2} \left(\sum_{i=1}^n \frac{( x_i - \hat{\mu})^2}{2\ \hat{\mu}^2 x_i}\right)^{-1}$$ where we plug in the MLE estimator $\mu$. The information matrix is positive-definite in these values.

It is hinted that $EX = \mu$, $E(1/X) = 1/\mu + 1/\lambda$, $VARX = \mu^3/\lambda$, $VAR1/X = \frac{1}{\mu \lambda} + \frac{2}{\lambda^2},$ and $COV(X,1/X) = -\mu/\lambda$.

I am not sure how to continue. I guess asymptotic normality of both follows by using CLT on $\hat{\mu}$ and then use the delta-method? The consistency of $\hat{\mu}$ follows from Chebyshev's inequality, but I have no clue how to show it for $\hat{\lambda}$.

Indeed my problem is that I am not sure how those hints of variances of $1/X$ and so on are supposed to come into play. Any answer or hint is appreciated.

| cite | improve this question | | | | |

$$\frac{n}{2} \left(\sum_{i=1}^n \frac{( x_i - \hat{\mu})^2}{2\ \hat{\mu}^2 x_i}\right)^{-1} = \hat{\mu}^2 \cdot \left(\frac 1n\sum_{i=1}^n \frac{( x_i - \hat{\mu})^2}{\ x_i}\right)^{-1} =$$

$$=\hat{\mu}^2 \cdot \left(\frac 1n\sum_{i=1}^n \left (x_i -2\hat {\mu} + \frac{\hat{\mu}^2}{x_i}\right) \right)^{-1}$$

So you are looking at sample means of an ergodic sample. Apply the probability limit and its properties (including the continuous mapping theorem), use the consistency for $\mu$, and Law of Large Numbers, as well as the hint about $E(1/X)$ (which importantly also implies that the expected value of the reciprocal exists in the first place).

These will lead you to consistency for $\lambda$.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.