Suppose $(X,Y)$ has the pdf

$$f_{\theta}(x,y)=e^{-(x/\theta+\theta y)}\mathbf1_{x>0,y>0}\quad,\,\theta>0$$

Density of the sample $(\mathbf X,\mathbf Y)=(X_i,Y_i)_{1\le i\le n}$ drawn from this population is therefore

\begin{align} g_{\theta}(\mathbf x,\mathbf y)&=\prod_{i=1}^n f_{\theta}(x_i,y_i) \\&=\exp\left[{-\sum_{i=1}^n\left(\frac{x_i}{\theta}+\theta y_i\right)}\right]\mathbf1_{x_1,\ldots,x_n,y_1,\ldots,y_n>0} \\&=\exp\left[-\frac{n\bar x}{\theta}-\theta n\bar y\right]\mathbf1_{x_{(1)},y_{(1)}>0}\quad,\,\theta>0 \end{align}

The maximum likelihood estimator of $\theta$ can be derived as

$$\hat\theta(\mathbf X,\mathbf Y)=\sqrt\frac{\overline X}{\overline Y}$$

I wish to know whether the limiting distribution of this MLE is normal or not.

It is clear that a sufficient statistic for $\theta$ based on the sample is $(\overline X,\overline Y)$.

Now I would have said that the MLE is asymptotically normal without a doubt if it was a member of the regular one-parameter exponential family. I don't think that is the case, partly because we have a two-dimensional sufficient statistic for a one-dimensional parameter (as in $N(\theta,\theta^2)$ distribution, for example).

Using the fact that $X$ and $Y$ are in fact independent Exponential variables, I can show that the exact distribution of $\hat\theta$ is such that

$$\frac{\hat\theta}{\theta}\stackrel{d}{=} \sqrt F\quad,\text{ where }F\sim F_{2n,2n}$$

I cannot possibly proceed to find the limiting distribution from here.

Instead I can argue by WLLN that $\overline X\stackrel{P}\longrightarrow\theta$ and $\overline Y\stackrel{P}\longrightarrow 1/\theta$, so that $\hat\theta\stackrel{P}\longrightarrow\theta$.

This tells me that $\hat\theta$ converges in distribution to $\theta$. But this does not come as a surprise, since $\hat\theta$ is a 'good' estimator of $\theta$. And this result is not strong enough to conclude whether something like $\sqrt n(\hat\theta-\theta)$ is asymptotically normal or not. I could not come up with a reasonable argument using CLT either.

So a question remains whether the parent distribution here satisfies the regularity conditions for the limiting distribution of MLE to be normal.

  • $\begingroup$ Empirically it seems to be very close to normal. You might find it easier to set $\theta$ to $1$ (it is only a scaling factor) and then consider whether the distribution of the square root of the ratio of sample means of i.i.d. exponential random variables is asymptotically normal. Using the delta method, this corresponds to the distribution of the ratio of sample means of i.i.d. exponential random variables being asymptotically normal. And that corresponds to the distribution of the ratio of two i.i.d. gamma random variables being asymptotically normal as the shape parameter increases. $\endgroup$
    – Henry
    Commented Apr 26, 2019 at 11:49
  • $\begingroup$ Asymptotic normality of MLEs has nothing to do with Exponential families. Intuitively, for asymptotic normality to hold you just need to make sure there is no chance that the solution will be near the boundary of the parameter space. $\endgroup$
    – whuber
    Commented Apr 26, 2019 at 13:38
  • $\begingroup$ @whuber As far as I know, pdfs that are members of the canonical exponential family almost always have MLEs that are asymptotically normal (not that it is due to the exp family). That is the connection I was trying to point out. $\endgroup$ Commented Apr 26, 2019 at 18:33
  • 1
    $\begingroup$ Right: but the connection is one way. The asymptotic results for MLE are far more general and therefore I was trying to suggest that looking in that general direction, rather than focusing on properties of Exponential families, might be a more fruitful inquiry. $\endgroup$
    – whuber
    Commented Apr 26, 2019 at 18:54
  • $\begingroup$ A proof using multivariate CLT and delta method is also possible as is done here. $\endgroup$ Commented Apr 7, 2020 at 19:45

1 Answer 1


A direct proof for asymptotic normality:

The log-likelihood here is

$$L = -\frac {n \bar x}{\theta} - \theta n \bar y$$

The first and second derivatives are

$$\frac {\partial L}{\partial \theta } = \frac {n \bar x}{\theta^2} - n\bar y,\;\;\;\frac {\partial^2 L}{\partial \theta^2 } = -\frac {2n \bar x}{\theta^3} $$

The MLE $\hat \theta_n$ satisfies

$$\frac {\partial L(\hat \theta_n)}{\partial \theta }=0$$

Applying a mean value expansion around the true value $\theta_0 $ we have

$$\frac {\partial L(\hat \theta_n)}{\partial \theta } = \frac {\partial L(\theta_0)}{\partial \theta } + \frac {\partial^2 L(\tilde \theta_n)}{\partial \theta^2 }(\hat \theta_n - \theta_0) =0$$

for some $\tilde \theta_n$ in between $\hat \theta_n$ and $\theta_0$. Re-arranging we have,

$$(\hat \theta_n - \theta_0) = -\left(\frac {\partial^2 L(\tilde \theta_n)}{\partial \theta^2 }\right)^{-1}\frac {\partial L(\theta_0)}{\partial \theta }$$

But in our single-parameter case, the inverse is just the reciprocal, so, inserting also the specific expressions of the derivatives,

$$(\hat \theta_n - \theta_0) = \frac {\tilde \theta^3_n}{2n\bar x}\left(\frac {n \bar x}{\theta^2_0} - n\bar y\right)$$

$$\implies \sqrt{n}(\hat \theta_n - \theta_0) = \frac {\tilde \theta^3_n}{2\bar x \theta_0^2}\sqrt{n}\cdot\left(\bar x - \theta_0^2\bar y \right)$$

$$\implies \sqrt{n}(\hat \theta_n - \theta_0) = \frac {\tilde \theta^3_n}{2\bar x \theta_0^2}\cdot\left (n^{-1/2}\sum_{i=1}^n(x_i-\theta_0^2 y_i)\right)$$

The variance of the sum is

$$\text{Var}\left(\sum_{i=1}^n(x_i-\theta_0^2 y_i)\right) = 2n\theta_0^2 $$

Manipulating the expression we can write, using $S_n$ for the sum of the i.i.d. elements,

$$\sqrt{n}(\hat \theta_n - \theta_0) = \left(\frac {\tilde \theta^3_n}{\sqrt{2}\bar x \theta_0}\right)\cdot\frac {\sum_{i=1}^n(x_i-\theta_0^2 y_i)}{\sqrt{n}\sqrt{2}\theta_0} $$

$$\sqrt{n}(\hat \theta_n - \theta_0) = \left(\frac {\tilde \theta^3_n}{\sqrt{2}\bar x \theta_0}\right)\cdot\frac {S_n}{\sqrt{\text{Var}(S_n)}}$$

More over, we have that $E(x_i-\theta_0^2 y_i) = 0$, so $E(S_n)=0$. So we have the subject matter of a classical CLT, and one can verify that the Lindeberg condition is satisfied. It follows that

$$\frac {S_n}{\sqrt{\text{Var}(S_n)}} \to_d N(0,1)$$

Due to the consistency of the estimator, we also have

$$\left(\frac {\tilde \theta^3_n}{\sqrt{2}\bar x \theta_0}\right) \to_p \frac{\theta_0}{\sqrt{2}}$$

and by Slutsky's Theorem we arrive at

$$\sqrt{n}(\hat \theta_n - \theta_0) \to_d N (0, \theta_0^2/2)$$

Nice. Double the information, half the variance (compared to the case where we would estimate $\theta_0$ based on a sample from a single random variable).

PS: The fact that in the above expressions $\theta_0$ appears in the denominator, points towards @whuber's comment that MLE's asymptotic normality needs the unknown parameter to be away from the boundary of the parameter space (in our case, away from zero).

  • $\begingroup$ Sorry for the late reply. All this time I was pondering whether this is a curved exponential family and so the MLE might behave differently. $\endgroup$ Commented Apr 29, 2019 at 22:00
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    $\begingroup$ @StubbornAtom Asymptotic normality is certainly lost when the parameter under estimation is on the boundary of the parameter (a quite intuitive result if you think about it). $\endgroup$ Commented Apr 30, 2019 at 0:43

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