Conditional Density Function Derivation

Question

Let $(\Omega, \mathcal{F},P)$ be a probability space and $X\colon\Omega \to \mathbb{R},Y \colon \Omega \to \mathbb{R}$ be continuous random variables (i.e. random variables which have a density function. I am assuming that this implies $P(X=x)=P(Y=y)=0~\forall x,y \in \mathbb{R}$).

According to Papoulis, the conditional distribution function $F_{X|Y} = P(X \leq x | Y = y)$ is defined by considering the probability $P(X \leq x | y \leq Y \leq y + \delta y)$ and taking the limit $\delta y\to 0$. However, I do not find the derivation given there rigorous.

It is easy to write:

$$P(X \leq x | y \leq Y \leq y + \delta y) = \frac{P( X \leq x, y \leq Y \leq y + \delta y)}{P(y \leq Y \leq y + \delta y)} = \frac{F_{X,Y}(x,y+\delta y) - F_{X,Y}(x,y)}{F_Y(y+\delta y) - F_Y(y)}$$ from definition of $F_{X,Y}$, $F_Y$ and the fact that the point probabilities are zero.

I am not sure how to evaulate the limit:

$$\lim_{\delta y \to 0} \frac{F_{X,Y}(x,y+\delta y) - F_{X,Y}(x,y)}{F_Y(y+\delta y) - F_Y(y)}.$$

I have tried using the L'Hopital rule as this limit is of the form $\frac{0}{0}$ but I am not sure if that is the right direction.

Any help is much appreciated.

EDIT: Papoulis obtains a formula for the density function by differentiating the term inside the limit. i.e., $$ f_{X|Y}(x,y) = \lim_{\delta y \to 0} \left(\frac{\partial F_{X,Y}(x,y+\delta y) - F_{X,Y}(x,y)}{\partial x}\frac{1}{F_Y(y+\delta y) - F_Y(y)}\right)$$

I believe I should have asked for the derivation of the density function as I am afraid that the Conditional Distribution functions does not have a neat expression.

Thanks, Phanindra

Answer 1

$P\left ( \{X \leq a\} \cap \{b \leq Y \leq b + \delta b\}\right )$ is the total probability mass in the (infinitely long) horizontal strip with NE corner $(a, b+\delta b)$ and SE corner $(a, b)$, or if we think of the joint density $f_{X, Y}(x,y)$ as defining a surface above the $x$-$y$ plane, then this probability can be thought of as the volume of a slice of width $\delta b$. For small values of $\delta b$ (when the slice is very thin), this volume is approximately the face area of the slice (the integral of the joint density $f_{X, Y}(x,b)$ from $x = -\infty$ to $x = a$) times the slice thickness $\delta b$. That is, $$ P\left (\{X \leq a\} \cap \{ b \leq Y \leq \delta b\} \right ) \approx \delta b \int_{-\infty}^a f_{X, Y}(x,b) \mathrm dx. $$ Since $P(b \leq Y \leq \delta b) \approx f_Y(b)\delta b$ for small values of $\delta b$, we get that $$ \begin{align*} P(X \leq a \mid b \leq Y \leq \delta b) &= \frac{P\left (\{X \leq a\} \cap \{ b \leq Y \leq \delta b\} \right )}{P(\{b \leq Y \leq \delta b\})}\\ &\approx \frac{\displaystyle \delta b \int_{-\infty}^a f_{X, Y}(x,b) \mathrm dx}{f_Y(b)\delta b}\\ &\approx \int_{-\infty}^a \frac{f_{X, Y}(x,b)}{f_Y(b)} \mathrm dx \end{align*} $$ The above should be thought of as a heuristic justification of the definition of conditional distribution function and density of $X$ conditioned on $Y = b$ as $$ \begin{align*} F_{X \mid Y}(a \mid Y = b) &= \int_{-\infty}^a \frac{f_{X, Y}(x,b)}{f_Y(b)} \mathrm dx\\ f_{X \mid Y}(a \mid Y = b) &= \frac{\partial}{\partial a}F_{X \mid Y}(a \mid Y = b) = \frac{f_{X, Y}(a,b)}{f_Y(b)} \end{align*} $$

Answer 2

$$ F_{X|Y}(x,y) = \lim_{\delta y \to 0} \left(\frac {F_{X,Y}(x,y+\delta y) - F_{X,Y}(x,y)}{F_Y(y+\delta y) - F_Y(y)}\right)$$ As this is of the form $\frac{0}{0}$, using the L'Hopital's rule, we get $$\lim_{\delta y \to 0} \left(\frac {\partial {F_{X,Y}(x,y+\delta y) - F_{X,Y}(x,y)}}{\partial \delta y} \frac{1}{\frac{\partial F_Y(y+\delta y) - F_Y(y)}{\partial \delta y}} \right)$$ which equals $$\frac{\partial {F_{X,Y}(x,y)}}{\partial y} \frac{1}{f_Y(y)}$$ which when differentiated w.r.t $x$ gives the expression for the density $$f_{X|Y}(x,y) := \frac{\partial F_{X|Y}(x,y)}{\partial x}=\frac{\partial^2 {F_{X,Y}(x,y)}}{\partial x\partial y} \frac{1}{f_Y(y)} := \frac{f_{X,Y}(x,y)}{f_Y(y)}$$

Hopefully, my assumption regarding the existence of partial derivative of $F_{X,Y}(x,y)$ such as $\frac{\partial {F_{X,Y}(x,y)}}{\partial y} $ is not wrong.