Conditioning on a random variable (1 Viewer)

InteGrand · Dec 26, 2016

He-Mann said:
$\noindent $Suppose that $X \sim U_{[0,1]}$ and $Y|(X = x) \sim U_{[0,x]}.$ The probability density function (PDF) of $Y$ can be found by conditioning on $X, \\ \mathbb{P}(Y = y) = \int_y^1 \mathbb{P}(Y = y|X = x)\mathbb{P}(X = x)dx. \\ $I understand how to identify the integral bounds. The random variables can model a slice of a random amount $(X)$ of cake, then using this slice, randomly slice it to obtain a final piece of cake $(Y|X)$. Looking at a numerical example, $y = 0.3$, we check what are the possible values that $x$ can take. Clearly, $x \leq 1.$ The lower bound for $x$ is $x \geq 0.3(=y)$, otherwise, if $x \leq 0.3$ (e.g. $ x = 0.2$, 20\% of the cake), then it is impossible that $y = 0.3$, 30\% of the cake (unless there is $\textit{magic}$ involved!). Does that make sense? Is there an easier way to think about this? \\\\ My question is, for two continuous random variables $X$ and $Y|X$, what is the general formula to find the PDF of $Y?$

$\noindent You could find the CDF of $Y$ using $F_{Y}(y) = \int_{\text{all }x} \mathbb{P}\left(Y\leq y\mid X = x \right)f_{X}(x)\, \mathrm{d}x$, where $f_{X}(x)$ is the density of $X$ and the $\mathbb{P}$ in the integral is the CDF of $Y\mid X=x$. Then differentiation would yield the PDF of $Y$. Or use $f_{Y}(y) = \int_{\text{all }x}f_{Y|X=x}(y)f_{X}(x)\,\mathrm{d}x$. (Here $f_{Y|X=x}(y)$ is the density of $Y|X=x$ evaluated at $y$.) The relevant bounds etc. would depend on the example at hand.$

He-Mann · Dec 26, 2016

InteGrand said:
$$\noindent You could find the CDF of $Y$ using $F_{Y}(y) = \int_{\text{all }x} \mathbb{P}\left(Y\leq y\mid X = x \right)f_{X}(x)\, \mathrm{d}x$, where $f_{X}(x)$ is the density of $X$ and the $\mathbb{P}$ in the integral is the CDF of $Y\mid X=x$. Then differentiation would yield the PDF of $Y$. The $\textbf{relevant bounds etc. would depend on the example at hand}.$

$\noindent Or use $f_{Y}(y) = \int_{\text{all }x} f_{Y|X=x}(y)f_{X}(x)\,\mathrm{d}x$. Here $f_{Y|X=x}(y)$ is the density of $Y|X=x$ and $f_{Y}(y)$ the density of $Y$.$

Everything makes sense now. Thank you heaps!

He-Mann · Dec 28, 2016

$\noindent\textbf{Problem}: Mrs. Jones has made a steak and kidney pie for her two sons. Eating more than a half of it will give indigestion to anyone. While she is away having tea with a neighbour, the older son helps himself to a piece of the pie. Then the younger son comes and has a piece of what is left by his brother. We assume that the size of each of the two pieces eaten by Mrs. Jones' sons is random and uniformly distributed over what is currently available. What is the expected size of the remaining piece given that neither son gets indigestion?$

$$\noindent \textbf{My solution}: Let $X \sim U_{[0,1]}$ and $Y|(X =x) \sim U_{[0,1-x]}$ be random variables representing the amount of pie eaten by the older and younger brother, respectively. We need to find the expected amount eaten: \\ $ \begin{align*} \mathbb{E}\left(X \leq \frac{1}{2}, Y \leq \frac{1}{2} \right) & = \int_0^{1/2} \int_0^{1/2} xy \mathbb{P}\left(X \leq \frac{1}{2}, Y \leq \frac{1}{2}\right)dydx\\ &= \int_0^{1/2} \int_0^{1/2} xy \left( \int_0^{1/2} \int_0^{y} \frac{1}{1-x}dxdy \right )dydx \\ &= \dots\end{align*} \\ $The second line is because of \\ $ \mathbb{P}(X = x, Y = y) = \mathbb{P}(Y = y | X = x) \mathbb{P}(X = x) = \frac{1}{1-x} \times 1.$ Then subtract the above computed expectation from 1 and multiply by 100 to yield percentage remaining pie.$

Is this correct?

InteGrand · Dec 28, 2016

He-Mann said:
$\noindent\textbf{Problem}: Mrs. Jones has made a steak and kidney pie for her two sons. Eating more than a half of it will give indigestion to anyone. While she is away having tea with a neighbour, the older son helps himself to a piece of the pie. Then the younger son comes and has a piece of what is left by his brother. We assume that the size of each of the two pieces eaten by Mrs. Jones' sons is random and uniformly distributed over what is currently available. What is the expected size of the remaining piece given that neither son gets indigestion?$

$$\noindent \textbf{My solution}: Let $X \sim U_{[0,1]}$ and $Y|(X =x) \sim U_{[0,1-x]}$ be random variables representing the amount of pie eaten by the older and younger brother, respectively. We need to find the expected amount eaten: \\ $ \begin{align*} \mathbb{E}\left(X \leq \frac{1}{2}, Y \leq \frac{1}{2} \right) & = \int_0^{1/2} \int_0^{1/2} xy \mathbb{P}\left(X \leq \frac{1}{2}, Y \leq \frac{1}{2}\right)dydx\\ &= \int_0^{1/2} \int_0^{1/2} xy \left( \int_0^{1/2} \int_0^{y} \frac{1}{1-x}dxdy \right )dydx \\ &= \dots\end{align*} \\ $The second line is because of \\ $ \mathbb{P}(X = x, Y = y) = \mathbb{P}(Y = y | X = x) \mathbb{P}(X = x) = \frac{1}{1-x} \times 1.$ Then subtract the above computed expectation from 1 and multiply by 100 to yield percentage remaining pie.$

Is this correct?

$\noindent Why do you have $xy$ in the integral? One possible approach is as follows.$

$\noindent Note that our sample space $\Omega$ is the part of the unit square below the line $y = 1-x$ in the $x$-$y$ plane. That is, the interior and boundary of the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$. For simplicity let $A$ denote the event that neither son gets indigestion, that is, $A$ is the event $\left\{X\leq \frac{1}{2}, Y\leq \frac{1}{2}\right\}$. So conditioning on $A$, our new sample space becomes the square (call it $S$) with lower-left vertex the origin and side lengths $\frac{1}{2}$, i.e. vertices $(0,0), \left(\frac{1}{2},0\right)$, $\left(\frac{1}{2},\frac{1}{2}\right), \left(0,\frac{1}{2}\right)$. We want to find $\mu:= \mathbb{E}\left[X+Y\mid A\right]$ (conditional expected value of total eaten proportion given no indigestion). Subtracting this from $1$ will give us the desired answer (as a ratio).$

$\noindent One way to find $\mu$ is to first find the conditional joint density function of $(X,Y)$ given $A$. If we find this (denote it as $f_{A}(x,y)$), then $\mu$ is given by $\mu = \iint_{S}(x+y) f_{A}(x,y)\,\mathrm{d}x\,\mathrm{d}y$. So now we'll just focus on finding this conditional joint density function $f_{A}(x,y)$.$

$\noindent The joint density is just going to be the second-order mixed partial derivative of the joint CDF. That is, $f_{A}(x,y) = \frac{\partial^{2}}{\partial x\partial y}\left(F_{A}(x,y)\right)$, where $F_{A}(x,y) = \mathbb{P}\left(X\leq x,Y \leq y\mid A\right)$, for $x,y\in S$ (conditional joint CDF). Denote $F(x,y)$ the unconditional joint CDF for $x,y\in \Omega$. Let $p:= \mathbb{P}\left(A\right)$ $\Bigg{(}$so $p = F\left(\frac{1}{2},\frac{1}{2}\right)$$\Bigg{)}$. Then for $x,y\in S$, we have$

$\begin{align*}F_{A}(x,y) &= \mathbb{P}\left(X\leq x, Y\leq y \mid A\right)\\ &= \frac{\mathbb{P}\left(X\leq x, Y\leq y, A\right)}{p} \\ &= \frac{\mathbb{P}\left(X\leq x, Y\leq y\right)}{p} \quad (\text{as }x,y\in S) \\ &= \frac{F(x,y)}{p}.\end{align*}$

$\noindent Taking mixed second-order partials of both sides then yields $\boxed{f_{A}(x,y) = \frac{f(x,y)}{p}}$, where $f(x,y)$ is the unconditional joint density function. Note that the joint PDF is $f(x,y) = f_{Y|X = x}(y) f_{X}(x) = \frac{1}{1-x}$, for $x,y \in \Omega$. (We can use this to calculate $p$ also.)$

He-Mann · Dec 28, 2016

Thanks for that. Not sure what I was doing in my post; forgot the basics of expectation. I put xy because I was confused between the ideas of counting (adding is AND, multiplying is OR) and common sense.

Whenever I'm doing maths, my common sense just gets ignored (usually for probability). I get caught up with all the notation and stuff, so I forget the fundamental stuff like common sense. It's probably that I'm not aware of what I'm doing (due to autopilot) and this is probably exacerbated by a fragmented and methodical understanding of counting.

leehuan · Jan 7, 2017

What course is this He-Mann

He-Mann · Jan 8, 2017

leehuan said:
What course is this He-Mann

The question is from one of MATH3901's class tests and can be done with MATH2901 knowledge. Just prepping for MATH3901 next sem and the future by re-learning probability and stats again with: https://www.youtube.com/playlist?list=PL2SOU6wwxB0uwwH80KTQ6ht66KWxbzTIo (Harvard's Statistics 110).

InteGrand · Jan 8, 2017

Incidentally, here's the course song for Harvard's Stat 110:

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Conditioning on a random variable (1 Viewer)

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