$$\frac{e^{-(y-θx)^2/2x^2) -x/λ}}{(λ\sqrt{2πx^2})}$$

This is the joint distribution function.

a)I have to find the marginal function of $X$ and $Y|X$. Now the $X\sim \text{Exp}(1/λ)$ and $Y|X\sim N(θx;x^2)$

b)Then, find the MLE for $λ$ and $θ$.
For $λ$ we have: $\cal{L}= \prod \left[e^{(\frac{-x}{\lambda})}/\lambda\right] =[1/λ]^n e^{(-∑ 〖x/λ〗)}$

$\log(\cal{L})= -\log\lambda^n-(\sum x)/\lambda $

$\frac{dl}{d\lambda}= -n/\lambda+\sum x/\lambda^2 =0$

$\hat \lambda =\sum x/n $

and for $θ$: $\cal{L}=\prod [\frac{1}{\sqrt{2\pi}}\frac{1}{x} e^{(〖-(y-\theta x)〗^2/ 2x^2 )} ] $

$=1/\prod x\cdot e^{\sum \frac{〖-(y-θx)〗^2}{2x^2}}$

$\log⁡(\cal{L})=-\sum \log ⁡x - \sum y^2/〖2x〗^2 + \sum \theta y/x- \theta^2/2$

I derive $\theta$ and resolve

$\hat\theta = \sum y/x$

c) Are they unbiased?

$E(\hat{\lambda})=\lambda$. For $\theta$ do I have to use the law of iterated expectation?

I have the same question for the variance of $\theta$. We know that for large n we have asymptotically unbiased mle.

d)Exact distribution of $\lambda$ and $\theta$? Asymptotically the distribution for $\theta$ and $\lambda$ is normal, but is it possible to find the exact distribution for theta? (for $\lambda$ it is a $\text{gamma}(n,n/\lambda)$, right?)

I have to try with the transformation law, given that x is a function of y. I have many doubts here!

  • 4
    $\begingroup$ See the self-study tag wiki if applicable. $\endgroup$ Commented Jun 4, 2014 at 23:34
  • $\begingroup$ Affter you read the tag wiki Nick linked to, please modify your question accordingly. $\endgroup$
    – Glen_b
    Commented Jun 4, 2014 at 23:53
  • $\begingroup$ For (c), just perform the integration. For (d), change variables from $(x,y)$ to $(x,y/x)$. $\endgroup$
    – whuber
    Commented Jun 5, 2014 at 12:50
  • $\begingroup$ For d) I do not understand how I should proceed $\endgroup$
    – user46784
    Commented Jun 5, 2014 at 13:36
  • $\begingroup$ I think there is a mistake in the mle for the conditional normal distribution. It should read $\hat{\theta}=\bar{y}/x $ cc @whuber $\endgroup$
    – JohnK
    Commented Jun 5, 2014 at 14:34

1 Answer 1


In a reply concerning interpreting probability densities I have argued for the merits of including the differential terms ($dx$ and $dy$ in this case) in the PDF, so let's write

$$f(x,y; \theta,\lambda) = \frac{1}{\lambda x \sqrt{2\pi}} \exp\left(-\frac{(y-x\theta)^2}{2x^2} - \frac{x}{\lambda}\right)\,dx\,dy.$$

In (d), attention is focused on $\hat{\theta}$ which appears to be a multiple of $\sum_i y_i/x_i$ where each $(x_i,y_i)$ is an independent realization of the bivariate random variable described by $f$. To tackle this, let's consider the distribution of $Z=Y/X$--and then we will later have to sum a sequence of independent versions of this variable in order to determine the distribution of $\hat\theta$.

In contemplating the change of variable $(X,Y)\to (X,Z) = (X,Y/X)$, we recognize that the non-negativity of $X$ assures this induces an order-preserving, one-to-one correspondence between $Y$ and $Z$ for each $X$. Therefore all we need to do is change the variable in the integrand, writing $y=x z$:

$$f(x,x z; \theta,\lambda) = \frac{1}{\lambda x \sqrt{2\pi}} \exp\left(-\frac{(x z-x\theta)^2}{2x^2} - \frac{x}{\lambda}\right)\,dx\,d(x z).$$

From the product rule $d(x z) = z dx + x dz$, and understanding the combination $dx\, d(x z)$ in the sense of differential forms $dx \wedge d(x z)$, we mechanically obtain

$$dx\, d(x z) = dx \wedge d(x z) = dx \wedge (z dx + x dz) = z dx \wedge dx + x dx \wedge dz = x\,dx\,dz.$$

(This is the easy way to obtain the Jacobian determinant of the transformation.)

Therefore the distribution of $(X,Z)$ has PDF

$$\eqalign{ g(x,z;\theta,\lambda) = f(x, x z; \theta, \lambda) &= \frac{1}{\lambda x \sqrt{2\pi}} \exp\left(-\frac{(x z-x\theta)^2}{2x^2} - \frac{x}{\lambda}\right)\,x dx\,dz \\ &= \frac{1}{\lambda} \exp\left( - \frac{x}{\lambda}\right)\frac{1}{\sqrt{2\pi}}\,dx\ \exp\left(-\frac{( z-\theta)^2}{2}\right) \,dz. }$$

Because this has separated into a product of a PDF for $X$ and a PDF for $Z$, we see without any further effort that (1) $X$ and $Z$ are independent and (2) $Z$ has a Normal$(\theta,1)$ distribution.

Finding the joint distribution of $(\hat{\lambda}, \hat\theta)$, which are easily expressed in terms of $X$ and $Z$, is now straightforward.

  • $\begingroup$ I can see the virtues of forcing the differentials to be more explicit, particularly as an aid for people not used to the idea of a probability density. But it does mean the notation for $f$ no longer matches the standard notation found in textbooks, and would cause raised eyebrows if handed in as part of an assignment! If somebody wanted to keep the mechanics of this approach without abandoning convention, would it suffice to replace e.g. $f$ by $dF$ and $g$ by $dG$ throughout? $\endgroup$
    – Silverfish
    Commented Nov 19, 2014 at 23:13
  • $\begingroup$ @Silver Your suggestion would destroy most of the computational advantages. I am using standard notation; it should be recognizable by any first-year math graduate student. It just hasn't filtered down from mathematics into statistics. I don't blame statisticians: despite its suitability to relativity and E&M, it took the better part of the 20th century for physicists to recognize its merits! One way to dodge the issue is to do the calculations however you wish and then announce the results to the world (in your paper or whatever): let the readers check it themselves if they want :-). $\endgroup$
    – whuber
    Commented Nov 19, 2014 at 23:21
  • $\begingroup$ The differential forms and their wedge products are all perfectly standard notation, at least among the mathematically inclined - but writing them as part of the PDF isn't standard. It seems to me all the interesting mechanics are happening on the right hand side. So rather than try to redefine the PDF, why not just call the left hand side something else? In particular, why not $dF$? I might not be seeing something, but it seems to me all the useful manipulation on the right side would proceed as before. $\endgroup$
    – Silverfish
    Commented Nov 19, 2014 at 23:36
  • $\begingroup$ @Silver I see better what you mean now. That makes sense and it's pretty close to the practice I have personally adopted. One has to be careful because expressions like "$dF$" can also be interpreted as Lebesgue-Stieltjes measures rather than as differential forms (although clearly the two are closely allied). All in all, though, explicitly writing out the differentials can be very helpful in avoiding common mistakes when working with PDFs. It's something I would seriously consider teaching in place of the usual Jacobian formulas. $\endgroup$
    – whuber
    Commented Nov 19, 2014 at 23:40

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