We have a joint probability distribution $f_{X,Y}(x,y)=\frac{1}{10}$, defined over the domain $(x,y)\in[-1,1]\times[-2,2]\cup[1,2]\times[-1,1]$. From this, I need to find the conditional PDFs $f_{X|Y}(x|y)$ and $f_{Y|X}(y|x)$.

I have figured out the marginal PDFs,

\begin{align*} f_X(x)&=\left\{\begin{array}{lr}\frac{2}{5},&-1\leq x\leq 1\\ \frac{1}{5},&1<x\leq2\end{array}\right.\\ \\ f_Y(y)&=\left\{\begin{array}{lr}\frac{2}{10},&-2\leq y<-1\\\frac{3}{10},&-1\leq y\leq 1\\ \frac{2}{10},&1<y\leq 2\end{array}\right. \end{align*}

However, am stuck on how to find the conditionals, due to the various domain constraints. I thought about splitting it up into two/three areas, and then finding $f_{X|Y\in a}(x)$, but I wasn't sure how to move from this partition into a single function for $X|Y$ and $Y|X$, and every other idea I think of won't normalise.

Note: I need to eventually find the variances of $X$ and $Y$ using the the expectation and variance of $Y|X$ and $X|Y$, I'm assuming with the law of total variance? If the functions stay split up, I'm assuming I just evaluate each section individually and then add them up?

  • $\begingroup$ By definition, conditional pdf are simply: $f_{X\mid Y}(x\mid y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}$ and $f_{Y\mid X}(y\mid x)=\frac{f_{X,Y}(x,y)}{f_X(x)}$ $\endgroup$ – Graham Kemp Oct 2 '14 at 7:42
  • $\begingroup$ I tried doing that directly, the issue is that $f_Y(y)$ and $f_X(x)$ are different depending on where my $y$ and $x$ are. But, if I then define my conditional PDFs based on the domains of $x$ and $y$, I end up with PDFs that won't normalise. $\endgroup$ – keefles Oct 2 '14 at 8:05
  • $\begingroup$ Why won't it normalise for you? $\endgroup$ – Graham Kemp Oct 2 '14 at 8:43
  • $\begingroup$ I'd assume because I'm doing something wrong - but with each method I tried, the probabilities would not add to one (generally going over one). $\endgroup$ – keefles Oct 2 '14 at 12:12

$\begin{align} f_{X,Y}(x,y) & =\begin{cases}1/10 & : (x,y)\in [-1,1]\times[-2,2]\cup[1,2]\times[-1,1] \\ 0 & : \text{elsewhere} \end{cases} \\[3ex] f_X(x) & =\begin{cases}\int_{-2}^2 1/10 \operatorname d y & : x\in [-1,1] \\ \int_{-1}^{1} 1/10 \operatorname d y & : x\in(1, 2] \\ 0 & : \text{elsewhere} \end{cases} \\[1ex] & =\begin{cases}2/5 & : x\in[-1,1] \\ 1/5 & : x\in (1,2] \\ 0 & : \text{elsewhere} \end{cases} \\[3ex] f_{Y\mid X}(y\mid x) & =\frac{f_{X,Y}(x,y)}{f_X(x)} \\[1ex] & =\begin{cases}1/4 & : x\in[-1,1], y\in[-2,2] \\ 1/2 & : x\in (1,2],y\in[-1,1] \\ 0 & : \text{elsewhere} \end{cases} \\[3ex] \mathsf E_{Y\mid X}[Y\mid X] & = \begin{cases} \int_{-2}^{2} y /4 \operatorname d y & : x\in[-1,1]\\\int_{-1}^{1} y /2\operatorname d y & : x\in(1,2]\\ 0 & : \text{elsewhere} \end{cases} \\[1ex] & = 0 \\[3ex] \mathsf E_{Y\mid X}[Y^2\mid X] & = \begin{cases} \int_{-2}^{2} y^2 /4 \operatorname d y & : x\in[-1,1]\\\int_{-1}^{1} y^2 /2\operatorname d y & : x\in(1,2]\\ 0 & : \text{elsewhere} \end{cases} \\[1ex] & = \begin{cases}4/3 & : x\in[-1,1] \\ 1/3 & : x\in (1,2]\\ 0 & : \text{elsewhere} \end{cases} \\[3ex] \mathsf {Var}[Y] & = \mathsf E_X[\mathsf E_{Y\mid X}[Y^2\mid X]]-\mathsf E_X[\mathsf E_{Y\mid X}[Y\mid X]]^2 \\[1ex] & = \int_{-1}^1 4/3\times 2/5 \operatorname d x + \int_{1}^2 1/3\times 1/5\operatorname d x \\[1ex] & = \frac{17}{15} \end{align}$


To verify that the conditional probability density functions ensure that they (1) are strictly increasing and (2) the integral over the conditional domain is equal to 1.

Our $f_{Y\mid X}$ does not decrease when progressing along the interval of $Y$ for any fixed value of $X$.  It is in fact constant for each possible value of $X$ (though a different constant for different values).

$\begin{align} f_{Y\mid X}(y\mid x) & =\begin{cases}1/4 & : x\in[-1,1], y\in[-2,2] \\ 1/2 & : x\in (1,2],y\in[-1,1] \\ 0 & : \text{elsewhere} \end{cases} \end{align}$

Next we integrate with respect to $y$ over the entire conditioned domain for all given values of $X$.   Thus verifying that the conditional measure is $1$ for all possible values of $X$ (and $0$ elsewhere).   That is, it's the indicator function for the domain of $X$; as it should be. (We have ensured that: $\mathsf P(Y\in \Omega_Y \mid X=x) = \operatorname{\bf 1}_{\Omega_X}(x)$)

$\begin{align}\int_{\Omega_Y\mid X=x} f_{Y\mid X}(y\mid x)\operatorname d y & = \begin{cases}\int_{-2}^2 1/4\operatorname d y & : x\in [-1,1] \\ \int_{-1}^1 1/2 \operatorname d y & : x\in (1,2]\\0 & :\text{elsewhere}\end{cases} \\[1ex] & = \begin{cases}1 & : x\in[-1,1]\\ 1 & : x\in(1,2]\\ 0 & : \text{elsewhere}\end{cases} \\[1ex] & = \begin{cases}1 & : x\in[-1,2]\\ 0 & : \text{elsewhere}\end{cases} \\[1ex] & = \operatorname{\bf 1}_{[-1,2]}(x)\end{align}$

Thus this conditional function is indeed a conditional probability density measure.

  • $\begingroup$ That's one of the various things I tried - the issue is, $f_{Y|X}$ doesn't normalise as you've defined it there, as integrating over the space actually gives you 2. $\endgroup$ – keefles Oct 2 '14 at 12:18
  • $\begingroup$ @keefles, How did you get $2$? Wait, have you been trying to add the pieces? It's a conditional function; you integrate over the domain of $Y$ when given $X=x$, which should result in an indicator function for $X$. $\endgroup$ – Graham Kemp Oct 2 '14 at 12:50
  • $\begingroup$ In your edit, I'm confused how you've gotten 1. Both pieces come to 1, which would be fine if they were defined as two separate functions, but they aren't. Shouldn't the function add each separate area to 1? Also, what's an indicator function? I've never heard the term before, and googling it is somewhat confusing... Sorry for my being a pain - this is meant to be an easy question, and I normally can do the more difficult ones, something's just not clicking for me. >.<; $\endgroup$ – keefles Oct 3 '14 at 0:19
  • $\begingroup$ @keefles If you think of the join pdf as a measure of distribution over the area of the block, then the conditional pdf measures the distributions over lengths across the cross sections for given values. So you just need to integrate across those cross section and ensure it always comes to 1 for any particular value of the conditioning variable. (That is any value at every instance, not every value all at once.) $\endgroup$ – Graham Kemp Oct 3 '14 at 0:43
  • $\begingroup$ Eg: Pick one such value, $X=0.2$ say. For this all values of $Y$ fall into the range $[-2,2]$ so we need $\mathsf P(-2\leq Y\leq 2\mid X=0.2)=1$. We test $\int_{-2}^2 f_{Y\mid X}(y\mid 0.2)\operatorname d y$ and see if it does. Since $f_{Y\mid X}(y\mid 0.2)=(1/4) \operatorname{\bf 1}_{[-2,2]}(y)$ we find it does. $\endgroup$ – Graham Kemp Oct 3 '14 at 0:50

This is the kind of question that is easy to do (and almost trivially so in this case) if you draw a diagram and think geometrically, visualizing the joint density as an object (of volume $1$) sitting on the $x$-$y$ plane. In this instance, this solid is the union of two prisms with rectangular bases, a large rectangle (of area $8$) whose opposite corners are at $(-1,-2)$ and $(1,2)$ and a smaller rectangle (of area $2$) whose opposite corners are at $(1,-1)$ and $(2,1)$. In addition to this visualization of the joint density as a solid, the tools you need are the following.

  1. If $g(x)$ is a function that takes on nonnegative values only and $\int_{-\infty}^\infty g(x)\,\mathrm dx$ has finite value (call this value $A$; we call it the area under the curve), then $g(x)/A$ is a valid probability density function.

  2. If $f_{X,Y}(x,y)$ is a joint probability density function, then the value of the marginal density of $X$ at any (fixed) point $x_0$ is the area of the cross-section of the pdf solid that you see if the pdf solid is sliced by a vertical plane through the line $x=x_0$. That is, we get the area under the curve of the function $f_{X,Y}(x_0,y)$ which is a function of $y$ (remember that $x_0$ is just a fixed constant value).

  3. The conditional density of $Y$ given that $x = x_0$ is proportional to the function $f_{X,Y}(x_0, y)$ regarded as a function of $y$ alone. This function is nonnegative, obviously, and we can make it into a valid density by dividing by the "area under the curve" as described in Item 1. Item 2 assures us that the area under the curve is just $f_X(x_0)$ leading to the formula beloved by writers of textbooks on probability: $$f_{Y\mid X}(y \mid X=x_0) = \frac{f_{X,Y}(x_0, y)}{f_X(x_0)}.$$

All this is, of course, fine and dandy but blindly applying formulas is not needed in this simple case; just visualization!

  • For any fixed $x_0 \in [-1,1]$, the cross-section $f_{X,Y}(x_0, y)$ is a rectangle whose base extends from $y=-2$ to $y = +2$. Similarly, for any fixed $x_0 \in (1,2]$, the cross-section $f_{X,Y}(x_0, y)$ is a rectangle whose base extends fron $y=-1$ to $y=+1$. Hence,

For each $x_0 \in [-1,1]$, the conditional distribution of $Y$ given that $X = x_0$ is $U[-2,2]$.
For each $x_0 \in (1,2]$, the conditional distribution of $Y$ given that $X = x_0$ is $U[-1,1]$.

Note that we didn't really need to explicitly compute the value of $f_X(x_0)$ in order to determine the conditional density of $Y$.

The conditional mean of $Y$ given $X = x_0$ is trivial to obtain. We have

For each $x_0 \in [-1,2]$, $E[Y\mid X = x_0] = 0$.

The conditional variance of $Y$ given $X=X_0$ is also easy to write down if one remembers the formula $(b-a)^2/12$ or, if not, since $E[Y \mid X = x_0] = 0$, it is easy to compute $\operatorname{var}(Y\mid X = x_0) = E[Y^2\mid X = x_0]$.

For each $x_0 \in [-1,1]$, $\displaystyle \operatorname{var}(Y\mid X = x_0) = \frac{4^2}{12} = \frac 43$.
For each $x_0 \in (1,2]$, $\displaystyle\operatorname{var}(Y\mid X = x_0) = \frac{2^2}{12} = \frac 13$.


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