Interpreting Bayes' thereom with density functions

I'm studying about Bayes' theorem and according to the theorem:

$$P(A\mid B) = \frac{P(B\mid A)P(A)}{P(B)}$$

This I can understand where it comes from etc. But if I use Bayes' theorem on density functions:

$$f_{X|Y}(x\mid y) = \frac{f_{Y|X}(y\mid x)f_X(x)}{f_Y(y)}$$

This makes me raise some questions. What does this mean? Why is it true? Is there a proof for this?

nbro

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asked Dec 4 '13 at 16:16

jjepsuomi

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$\begingroup$ I wouldn't use the same symbol, $f$ to represent several different functions. Notice that one writes things like $\Pr(X\le x)$, distinguishing between the random variable (capital) $X$ and the number (lower-case) $x$, and similarly $f_X(3)$ is the value of the density function of the random variable (capital) $X$ at $3$, whereas $f_Y(3)$ is the value of the density function of the random variable (capital) $Y$ at $3$. If you write $f(x)$, it should mean the value of the function called $f$ at the number $x$, and $f(y)$ should mean the value of the SAME function $f$ at $y$. $\endgroup$ – Michael Hardy Dec 4 '13 at 16:27
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$\begingroup$ Try to express $P(X \in A | Y = y)$ and define this by $\lim_{\epsilon \to 0} P(X \in A | Y \in B_\epsilon) $ where $B_\epsilon = [y-\epsilon, y+\epsilon]$ $\endgroup$ – Petite Etincelle Dec 4 '13 at 18:03

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