This is my first question in Mathematics on Stack Exchange. Please forgive that this is a none sense question…
Question
I'd like to know a simple form of the division of popular probability density functions (PDF).
There is the conjugate prior with which the product can be represented with simple form. Is there such kind of notion about division of PDFs?
If possible, I also hope to know the interpretation of the division of two PDFs.
Background
I met the division of Multinomial distribution when I employed Multinomial distribution as a Bayesian prior P(H) and P(E).
$$
P(H\mid E) = \frac{P(E\mid H) \cdot P(H)}{P(E)}
$$
Because $P(E\mid H)$ is given, I have to compute just $P(H)/P(E)$.
I know that people usually employ below style formula. Then, the division of PDF is not necessary.
$$
P(M \mid E) = \frac{P(E \mid M)}{\sum_m {P(E \mid M_m) P(M_m)}} \cdot P(M)
$$
However, I think if $P(H)/P(E)$ can be represented with simple form, the calculation of Bayesian probability should be easier. Thanks.
Best Answer
Yes! Let $X,Y$ are two real valued random variables whose common density $f_{X,Y}(x,y)$ exists. The marginal densities, then, are $$f_X(x)=\int_{\mathbb R}f_{X,Y}(x,y)\ dx \ \ \text{ and } \ \ f_Y(y)=\int_{\mathbb R}f_{X,Y}(x,y)\ dy.$$ The conditional densities are defined as follows $$f_{X|Y=y}(x)=\frac{f_{X,Y}(x,y)}{f_Y(y)} \ \ \text{ and } \ \ f_{Y|X=x}(y)=\frac{f_{X,Y}(x,y)}{f_X(x)}.$$ (If both the nominator and the denominator are zeros then, by definition, let the quotient be zero.)
Let $H', E'$ be (measurable) sets of reals. Accordingly, let's define the following events $$H=\{X\in H'\} \text{ and }E=\{Y\in E'\}.$$ Then $$P(H|E)=\frac{P(H\cap E)}{P(E)}=\frac{\iint_{H'\times E'}f_{X,Y}(x,y)dxdy}{\int_{E'}f_Y(y)dy}=\frac{\int_{E'}f_{Y}(y)\left[\int_{H'}f_{X|Y=y}(x)dx\right]dy}{\int_{E'}f_Y(y)dy}$$ or $$P(H|E)=\frac{P(H\cap E)}{P(E)}=\frac{P(E|H)P(H)}{P(E)}=P(E|H)\frac{P(H)}{P(E)}=\frac{\int_{H'}f_X(x)dx}{\int_{E'}f_Y(y)dy}.$$
Now, let's see what further help is needed...