2
$\begingroup$

So I have this question where you are given two random variables, $X$ and $Y$. $X$ is a continuous random variable (represented as a mean) with a distribution of $Exp(1)$ (exponential with $\lambda = 1$) and $Y$ is a discrete random variable (represented as the actual probability) with the distribution of $Pois(\lambda)$ and the parameter is $X$. I am asked to find the joint probability distribution.

The solution suggests that $f_{x, y} (x, y) = p_{y | x} (y | x)$ but I am not sure how to got to this or why. I do not even know why they use the discrete formula for this or why they even used a conditional. After, they multiple the two distribution formulae for each variable. I am not sure why this is either.

$\endgroup$
5
  • $\begingroup$ Can you share the source, page number etc.? $\endgroup$
    – gunes
    Nov 28 '19 at 9:27
  • $\begingroup$ @gunes It's kind of a University lecture example and I am not sure if I can share it. However, I do believe I solved it but I would appreciate it if someone could confirm it. $\endgroup$ Nov 28 '19 at 10:17
  • 1
    $\begingroup$ I can’t say more without explicitly seeing the solution, however I don’t think $f(x,y)=p(x|y)$. And, in your first paragraf you haven’t clearly mentioned about the dependence between the RVs. $\endgroup$
    – gunes
    Nov 28 '19 at 10:22
  • $\begingroup$ $X$ is the mean of something and is distributed as $Exp(1)$ and Y is the actual value (i.e. not the mean) with parameter $X = x$ such that $Y | X = x$ is distributed as $Pois(X = x)$. I'll give my own answer right now and hopefully, you can confirm it. $\endgroup$ Nov 28 '19 at 10:25
  • $\begingroup$ @gunes Posted my own answer, please feel free to confirm or comment about it. Thanks. $\endgroup$ Nov 28 '19 at 10:36
2
$\begingroup$

Let me clear the notation. $X\sim\exp(\lambda=1)$, and given $X$, $Y$ is Poisson distributed with $\lambda=X$. By the definition of conditional probability, we have:$$f_{X,Y}(x,y)=f_{Y|X}(y|x)f_X(x)$$

where $$f_{Y|X}(y|x)=e^{-x}\frac{x^y}{y!},f_X(x)=e^{-x} $$ when $x> 0, y\in \mathbb{Z^+}$.

So, your answer is correct but $f_{x,y}(x,y)\neq p_{y|x}(y|x)$

$\endgroup$
10
  • 1
    $\begingroup$ It's given in the question. $Y$ is Poisson distributed with mean $X$ defines $f_{Y|X}(y|x)$, not $f_Y(y)$. $\endgroup$
    – gunes
    Nov 28 '19 at 17:28
  • 1
    $\begingroup$ Notations differ in different textbooks. $f$ letter is not supposed to represent continuous distributions alone. Some use $p$ for both; some use $f$ for both, or as yours $f$ for cont, $p$ for discrete. $\endgroup$
    – gunes
    Nov 28 '19 at 17:57
  • 1
    $\begingroup$ Yes, $Y$ is discrete. $\endgroup$
    – gunes
    Nov 28 '19 at 18:00
  • 1
    $\begingroup$ Yes, because $X$ is exponentially distributed, which is a continuous RV. $\endgroup$
    – gunes
    Nov 28 '19 at 18:05
  • 1
    $\begingroup$ Yes, check out mixed density: en.wikipedia.org/wiki/Joint_probability_distribution#/… $\endgroup$
    – gunes
    Nov 28 '19 at 18:09
1
$\begingroup$

The $Y$ with parameter $X$ just means: $Y | X = x$ (where $x$ is constant). $Y$ is distributed as $Pois(X = x)$.

Now, to find the joint probability density function of the two variables, you can use $p_{y | x} (y | x)$ since $Y$ is dependent on a constant $X = x$ value and the joint conditional probability has its second variable as a constant. Therefore, $f_{x, y} (x, y) = p_{y | x} (y | x)$. Now, since $x$ is constant, to find the probability for some $y$ given $x$, it is simply the probability of the $y$ times the probability of $x$ or mathematically: $= \frac{e^{-x}x^y}{y!}\cdot e^{-x}$ for $x > 0$ and $y = 0, 1, 2, \dots$ (all from the pdfs of the respective distributions).

The other way i.e. $p_{x | y} (x | y)$ is harder to get (or not possible) since you would need a random variable where $y$ is constant, but for this question, there is none.

$\endgroup$
1
  • 2
    $\begingroup$ I think you need to append this text to your original question, instead of posting as an answer. $\endgroup$
    – gunes
    Nov 28 '19 at 12:03

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.