For conditional distribution $$f_{X|Y}(x|y) = \frac{f(x,y)}{f_Y(y)}$$

this is the basic definition I know about conditional distribution

Consider n + m trials having a common probability of success. Suppose, however, that this success probability is not fixed in advance but is chosen from a uniform (0, 1) population. What is the conditional distribution of the success probability given that then + m trials result inn successes?

But for this question

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why for this problem conditional formula looks like

$$f_{X|N}(x|n) = \frac{P(N=n|X=x)f_X(x)}{P(N=n)}$$????


1 Answer 1


You say you understand $$f_{X \mid Y} (x \mid y) = \frac{f(x,y)}{f_Y(y)}.$$ Then you should also understand $$f_{Y \mid X} (y \mid x) = \frac{f(x,y)}{f_X(x)}$$ which can be rearranged as $$f(x,y) = f_{Y \mid X}(y \mid x) f_X(x).$$ Plugging this last expression for $f(x,y)$ into the first equation above yields $$f_{X \mid Y} (x \mid y) = \frac{f_{Y \mid X}(y \mid x) f_X(x)}{f_Y(y)}.$$ This is sometimes called Bayes's rule.

  • $\begingroup$ RHS has a typo -- $X|Y$ should be $Y|X$... $\endgroup$
    – dnqxt
    May 15, 2019 at 6:04

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