4
$\begingroup$

How can we prove that a draw $(x,y)$ from the joint probability distribution of two random variables $X$ and $Y$ can be obtained by first generating a draw $x$ from the marginal probability distribution of $X$ and then a draw $y$ from the conditional probability distribution of $Y$ given $X=x$?
Is this a simple consequence of the definition of a conditional probability distribution?

$\endgroup$
1

1 Answer 1

5
+100
$\begingroup$

In short, drawing from the joint distribution is equivalent to first drawing from one marginal and second from the corresponding conditional.

To answer a side question from @statmerkur in the comments, drawing from a distribution $F$ means producing a realisation of a random variable with distribution $F$, either by inducing the phenomenon distributed as $F$ (e.g., the time to an electron emission) or by simulating it (e.g., taking $F^{-1}(U)$ with $U$ a uniform variate when the distribution is univariate).

Considering that$$\mathbb P(X\in A,\ Y\in B)=\mathbb P(X\in A)\mathbb P(Y\in B|X\in A)$$ $-$where $\mathbb P(\cdot)$ denotes the probability of the event between parentheses$-$, we have \begin{align} \mathbb P\{X\in (\epsilon,\epsilon+\text d\epsilon),\ Y\in (\eta,\eta+\text d\eta)\}&=\int_{\epsilon}^{\epsilon+\text d\epsilon} \int_{\eta}^{\eta+\text d\eta} f_{Y|X}(y|x)\,\text dy\,f_X(x)\,\text dx\\ &=\mathbb E^X[\mathbb I_{(\epsilon,\epsilon+\text d\epsilon)}(X)\times\mathbb E^{Y|X}\{\mathbb I_{(\eta,\eta+\text d\eta)}(Y)|X\}] \end{align} where [in reply to @singlemalt]

  1. $f_{Y|X}(\cdot|\cdot)$ denotes the probability density of the conditional distribution of $Y$ given $X$
  2. $\mathbb I_A(X)$ is the indicator function that $X\in A$, equal to either 0 or 1, with $$\mathbb P(X\in A)=\mathbb E^X[\mathbb I_A(X)]$$
  3. $\mathbb E$ is a blackboard-bold "E" that is often used for representing `expectations'
  4. the superscript $X$ is indicating that the expectation is wrt the distribution of the rv $X$,
  5. and the superscript $Y|X$ is indicating that the expectation is computed wrt the conditional distribution of the rv $Y$ given the rv $X$, meaning that $X$ is considered as fixed when computing this conditional expectation.

Therefore, if [one simulates] $X\sim f_X(x)$, with realisation $x$, and [then one simulates] $Y\sim f_{Y|X}(y|x)$, $$\mathbb I_{(\epsilon,\epsilon+\text d\epsilon)}(X)\times\mathbb I_{(\eta,\eta+\text d\eta)}(Y)$$ is an unbiased estimator of $$\mathbb E^{(X,Y)}[\mathbb I_{(\epsilon,\epsilon+\text d\epsilon)}(X)\times\mathbb I_{(\eta,\eta+\text d\eta)}(Y)]=\mathbb E^X[\mathbb I_{(\epsilon,\epsilon+\text d\epsilon)}(X)\times\mathbb E^{Y|X}\{\mathbb I_{(\eta,\eta+\text d\eta)}(Y)|X\}]$$ for all $(\epsilon,\eta)$. Letting $\text d\epsilon$ and $\text d\eta$ go to zero (0) allows one to conclude that the joint distribution of $(X,Y)$ is obtained through this decomposition marginal-then-conditional, thus that the simulation is producing a realisation from the correct (joint) distribution.

Take the example of a bivariate Normal $$(X,Y)^\prime\sim\mathcal N_2(0_2,\Sigma)\qquad \Sigma=\left(\begin{matrix}1 &\rho\cr\rho &1\end{matrix} \right)\qquad \rho\in(-1,1)$$ Then drawing $X$ from the marginal $\mathcal N(0,1)$ is equivalent to setting $X=\epsilon_x$ with $\epsilon_x\sim \mathcal N(0,1)$. Further, drawing $Y$ from the conditional $\mathcal N(\rho x,1-\rho^2)$ is equivalent to setting $$Y=\rho x +\sqrt{1-\rho^2}\epsilon_y\qquad\epsilon_y\sim\mathcal N(0,1)$$
Therefore $$ \left(\begin{matrix}X\cr Y\cr\end{matrix} \right)= \overbrace{\left(\begin{matrix}1 &0\cr \rho &\sqrt{1-\rho^2}\cr\end{matrix} \right)}^A \left(\begin{matrix}\epsilon_x\cr \epsilon_y\cr\end{matrix} \right) \sim \mathcal N_2(0_2,\Sigma)$$ since $$A A^\mathsf{T}=\Sigma$$ This implies that the draw is truly from the joint distribution.

$\endgroup$
1
  • $\begingroup$ Xi'an, what is the superscript $X$ on the exponential symbol? Does the thin hollow rectangle symbol represent identity? $\endgroup$ Aug 2, 2022 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.