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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.

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  • $\begingroup$ 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. $\endgroup$ – user12055579 Nov 28 '19 at 10:17
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    $\begingroup$ 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. $\endgroup$ – user12055579 Nov 28 '19 at 10:25
  • $\begingroup$ @gunes Posted my own answer, please feel free to confirm or comment about it. Thanks. $\endgroup$ – user12055579 Nov 28 '19 at 10:36
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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 $f_{x,y}(x,y)\neq p_{y|x}(y|x)$

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Yes, $Y$ is discrete. $\endgroup$ – gunes Nov 28 '19 at 18:00
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    $\begingroup$ Yes, because $X$ is exponentially distributed, which is a continuous RV. $\endgroup$ – gunes Nov 28 '19 at 18:05
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    $\begingroup$ Yes, check out mixed density: en.wikipedia.org/wiki/Joint_probability_distribution#/… $\endgroup$ – gunes Nov 28 '19 at 18:09
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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.

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    $\begingroup$ 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

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