I am so puzzled by this problem.

Given two variables $X_1$ and $X_2$, such that $X_i \sim \mathrm{Gam}(a_i, b)$, find the joint distribution of $X_1$ and $X_2$.

I understand how to proceed if variables are independent, but the general case is a bit unclear. Thanks for your help!

  • 3
    $\begingroup$ Just out of curiosity, is your statement of the question taken verbatim from the assignment? $\endgroup$ – cardinal Jan 17 '12 at 16:19
  • 1
    $\begingroup$ Yes, this is the only part that I could not understand. Basically I need to derive joint distribution of $Y_1$ $Y_2$ given transformation of the variables $Y_1=X_1+X_2$ and $Y_2=\frac{X_1}{X_1+X_2}$. I know how to proceed, but since I can't assume independence of $X_1$ and $X_2$, I am not sure how to get joint distribution of $X_1$ and $X_2$ in the first place. $\endgroup$ – notrockstar Jan 17 '12 at 17:41
  • 1
    $\begingroup$ It seems to me you have to assume independence to get anywhere, and furthermore, given the apparent level of the question (intro math/stat), it's entirely plausible that the question is implicitly assuming independence. Could you quote the entire question? $\endgroup$ – jbowman Jan 17 '12 at 18:01

As stated the problem does not make sense, because a joint distribution cannot be found from the marginal distributions! The only meaningful case (as an homework) is to assume independence. In which case the density of the joint distribution is obviously the product of both densities...

  • 2
    $\begingroup$ Perhaps more to the point: What do you mean by solvable since there is nothing much to solve here? Pick an arbitrary two-dimensional copula $C$, for example. Taking $C(G_1(x_1),G_2(x_2))$ seems to be an equally valid "solution" and, indeed, I believe every solution is of this form. The case of independence corresponds to $C(u,v) = uv$. $\endgroup$ – cardinal Jan 17 '12 at 14:42
  • 5
    $\begingroup$ +1. The fact that the joint distribution cannot be derived from the marginals is a very important remark! $\endgroup$ – Macro Jan 17 '12 at 14:50
  • 4
    $\begingroup$ @Macro (+1). This point cannot be overemphasized. It seems to be a common point of confusion among students, and, apparently, (in the OP's case) some teachers. :) $\endgroup$ – cardinal Jan 17 '12 at 15:28
  • 2
    $\begingroup$ @Xi'an, I am glad it didn't make sense to me either. Here is the full context of the problem: Basically I need to derive joint distribution of $Y_1$ $Y_2$ given transformation of the variables $Y_1=X_1+X_2$ and $Y_2=\frac{X_1}{X_1+X_2}$. I know how to proceed, but since I can't assume independence of $X_1$ and $X_2$, I am not sure how to get joint distribution of $X_1$ and $X_2$ in the first place. $\endgroup$ – notrockstar Jan 17 '12 at 17:44
  • 2
    $\begingroup$ I would suggest you solve the homework under this independence assumption since this is the only way you can solve it. Or at least the most likely missing item of information in the homework text. Stress very clearly from the start of the homework you are making this additional assumption and proceed under this assumption. $\endgroup$ – Xi'an Jan 17 '12 at 17:50

OP notrockstar knows the solution for the case when the random variables are independent but presumably cannot use it since a solution without the independence assumption is being sought. Perhaps the OP has posted only a simplified version of the question, and what has been left out makes a solution possible. For example, if $X_1$ and $X_2$ are given to be the times of the $k$-th and $(k+\ell)$-th arrivals in a Poisson process of intensity (arrival rate) $\lambda$, then these are Gamma random variables with order parameters $k$ and $k+\ell$ respectively. Furthermore, conditioned on $X_1 = x_1$, $X_2$ is a displaced Gamma random variable with order parameter $\ell$, that is, $X_2 = x_1 + Y$ where $Y$ is a Gamma random variable with order parameter $\ell$. Thus, $$\begin{align*} f_{X_1,X_2}(x_1,x_2) &= f_{X_2|X_1}(x_2|x_1)f_{X_1}(x_1)\\ &= \begin{cases} f_Y(x_2-x_1)f_{X_1}(x_1), & 0 < x_1 < x_2 < \infty,\\ 0, & \text{otherwise.} \end{cases} \end{align*}$$

In view of the additional information provided by the OP that what is really wanted is the joint distribution of $Y_1 = X_1 + X_2$ and $Y_2 = \frac{X_1}{X_1+X_2}$, maybe the problem is intended as drill in transformation of variables: can you express the joint density $f_{Y_1,Y_2}(y_1,y_2)$ in terms of the joint density $f_{X_1,X_2}(\cdot,\cdot)$ as $J(y_1,y_2)f_{X_1,X_2}(g_1(y_1,y_), g_2(y_1, y_2))$ with the Gamma functions thrown in as distractions, or merely as hints that $X_1, X_2 \in (0, \infty)$ to see if the students can deduce that $Y_2 \in (0,1)$.

This problem is readily solvable since it is easy to invert the transformation, find the Jacobian etc. At the end, one could say something like "If $X_1$, $X_2$ are assumed to be independent (this is not stated in the problem given) random variables with Gamma distributions, then the joint density $f_{X_1,X_2}(\cdot,\cdot)$ factors into the product of the marginal densities, and in this case, $f_{Y_1,Y_2}(y_1,y_2)$ equals "$\cdots$" possibly adding that $Y_1$ and $Y_2$ are obviously independent if they are (I don't believe they are but am willing to abide a proof that they are), or giving their marginal pdfs too etc.

In summary, $f_{Y_1,Y_2}(y_1,y_2)$ can be stated in terms of the joint density $f_{X_1,X_2}(\cdot,\cdot)$ without knowing the exact form of $f_{X_1,X_2}$ or the marginal densities of $X_1$ and $X_2$. The assumption that $X_1$, $X_2$ are independent can be used at the very end to say explicitly what $f_{Y_1,Y_2}(y_1,y_2)$ is; the Gammaity or independence of $X_1$ and $X_2$ is not needed or used at all in the earlier work, and indeed serves merely to clutter up the calculations without shedding much light on the matter.

  • $\begingroup$ I completely understand your explanation, however, I have no information about the relationship between $X_1$ and $X_2$ besides both being Gamma random variables . $\endgroup$ – notrockstar Jan 17 '12 at 17:46

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.