Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the probability density function of the random variable $X$ is $f_X(x)$ and the probability of set $A=\{x:a<X<b\}.$ How can we find the conditional probability density function $f_{X\mid A}(x)$?

My attempt:

When $x\notin A$, $f_{X\mid A}(x)=0.$

Correction according to comment When $x\in A$, $f_{X\mid A}(x)=\cfrac{f_X(y)}{P(A)}$ where $P(A)$ is the probability of even $A$.

This seems to give a valid probability distribution that sums to 1. But I am not sure if it is correct. Also is it a definition that I just wrote? Or can we derive it from some fundamentals?

Thanks a lot in advance.

share|cite|improve this question
No need to integrate in the numerator, only in the denominator: $$f_{X\mid A}(x\mid A) = f_{X\mid \{a<X<b\}}(x\mid a < X < b) = \begin{cases}\frac{f_X(x)}{\int_a^b f_X(x)\,\mathrm dx},&a < x < b,\\0,&\text{otherwise.}\end{cases}$$ – Dilip Sarwate Oct 10 '13 at 19:06
@DilipSarwate Thank you, I corrected accordingly and understand the mistake. – triomphe Oct 10 '13 at 19:40
up vote 2 down vote accepted

Or can we derive it from some fundamentals?

Of course we can. Recall that the density $f_X$ is uniquely defined (well, uniquely up to sets of zero Lebesgue measure) by the condition that, for every measurable bounded function $u$, $$ E[u(X)]=\int u(x)f_X(x)\mathrm dx. $$ Likewise, the conditional density $f_{X\mid A}$ of $X$ conditional on $A=[a\lt X\lt b]$, assuming that $P[A]\gt0$, is uniquely defined by the condition that, for every measurable bounded function $u$, $$ E[u(X)\mid A]=\int u(x)f_{X\mid A}(x)\mathrm dx. $$ The LHS is the ratio of $E[u(X);a\lt X\lt b]$ by $P[A]$. Now, $E[u(X);a\lt X\lt b]=E[v(X)]$ for $v:x\mapsto u(x)\mathbf 1_{a\lt x\lt b}$ hence $$ E[u(X);a\lt X\lt b]=\int v(x)f_X(x)\mathrm dx=\int u(x)f_X(x)\mathbf 1_{a\lt x\lt b}\mathrm dx. $$ Dividing by $P[A]$, this proves that $$ f_{X\mid A}(x)=\frac{\mathbf 1_{a\lt x\lt b}}{P[A]}f_X(x). $$ Note finally that $f_{X\mid A}\geqslant0$, as it should be for a density, and that, equally as it should be for a density, $$ \int f_{X\mid A}=\int\frac{\mathbf 1_{a\lt x\lt b}}{P[A]}f_X(x)\mathrm dx=\frac1{P[A]}\int_a^bf_X(x)\mathrm dx=1. $$

share|cite|improve this answer
thank you. Could you please explain the first claim, "density $f(x)$ is uniquely defined by....." Do you mean if we give expected value of just any bounded measurable function, we have the density? – triomphe Oct 11 '13 at 12:26
Yes, if the identity $E[u(X)]=\int u(x)f(x)\mathrm dx$ holds for every bounded measurable $u$, it implies that $f$ is the density of $X$. For example, $u=\mathbf 1_B$ shows that $P[X\in B]=\int_Bf$. – Did Oct 11 '13 at 21:08

You are almost correct: The conditional probability of X|A will be $f_{x|A}(x) = \frac{f_{X}(x)\textbf{1}_{A}}{\int_{a}^{b} f_{X}}$ so that you have a rescaled density.

This is derived from the definition of a conditional density, which can be derived from the definition of conditional probability: $\lim\limits_{\epsilon \longrightarrow 0} \frac{F_{X}(x+\epsilon)-F_{X}(x)}{\epsilon \int_{a}^{b} f_{X}}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.