5

Let $X$ be a continuous random variable with probability density function $f_X$. The way that I think about the meaning of $f_X$ is this: if $\Delta x$ is a small positive number then $$ P(X \in [x,x+ \Delta x]) \approx f_X(x) \,\Delta x. $$

Now suppose that $Y$ is also a continuous random variable. Is there a similar way to think about the conditional probability density function $f_{X \mid Y}(x \mid y)$? Notice that I cannot write $$ \tag{1} P(X \in [x,x + \Delta x] \mid Y = y) \approx f_{X \mid Y}(x \mid y) \, \Delta x $$ because the event $Y = y$ has probability $0$, and I can't condition on an event with probability $0$.

Question: Since equation (1) does not make sense, how should I think about $f_{X \mid Y}(x \mid y)$?

By the way, Bertsekas's book Introduction to Probability defines $f_{X\mid Y}$ like this: $$ f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}. $$ However, I do not find this formula to be intuitive, despite the formula's superficial similarity with the formula $$ P(A \mid B) = \frac{P(A \cap B)}{P(B)}. $$

Michael Hardy
  • 1
eternalGoldenBraid
  • 2,089
  • 1
    I think it would be better two write $f_{X,\mid,Y,=,y} (x).$ It's not until you understand that way of looking at this that you understand it correctly. $\qquad$ – Michael Hardy Jun 03 '18 at 01:18

1 Answers1

4

Let $A = \{X \in [x,x+\Delta x]\}$ and $B = \{Y \in [y,y+\Delta y]\}$.

By your reasoning

$$P(A|B) =\frac{P(A \cap B)}{P(B)} \approx \frac{f_{XY}(x,y) \Delta x \Delta y}{f_{Y}(y) \Delta y} = \frac{f_{XY}(x,y) }{f_{Y}(y)} \Delta x.$$

Now think of $P(X \in [x,x+\Delta x] | Y= y)$ (intuitively) as

$$P(X \in [x,x+\Delta x] | Y= y)= \lim_{\Delta y \to 0}P(A|B) \approx \frac{f_{XY}(x,y) }{f_{Y}(y)} \Delta x, $$

where $\displaystyle f_{X|Y}(x,y) = \frac{f_{XY}(x,y) }{f_{Y}(y)}.$

RRL
  • 90,707