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What is an elegant way of calculating this result (programmatically) in mathematica?

Let $X$ be a random variable with PDF

$$f_{X}(x) = \begin{cases} x/4, & \text{if $1<x\leq3$} \\ 0, & \text{otherwise} \end{cases}$$

Let $A$ be the event $\{X\geq2\}$

Find $f_{X|A}(x)$


Here is the answer done manually for convenience:

$$f_{X|A}(x) = \begin{cases} \frac{f_{X}(x)}{P(A)}, & \text{if $2\leq x\leq3$} \\ 0, & \text{otherwise} \end{cases}$$

$$P(A)=P(X\geq2)=\int_{2}^{3}{f_{X}(x) dx}=5/8$$

$$\implies f_{X|A}(x) = \begin{cases} \frac{2x}{5}, & \text{if $2\leq x\leq3$} \\ 0, & \text{otherwise} \end{cases}$$

I am looking for a programmatic solution that utilizes \[Conditioned] or something else as high-level.

TransformedDistribution and ProbabilityDistribution look like promising avenues, but I am unable to state my problem in terms of them.

I know I could write out the problem explicitly in low level code. But I am looking for an implicit, declarative implementation. Surely mathematica with all its high level abstraction power can do this?


here is a failed attempt

A = X >= 2;
conditionalDist = TransformedDistribution[X \[Conditioned] A, X \[Distributed]ProbabilityDistribution[x/4, {x, 1, 3}] ] ;
PDF[conditionalDist, y]
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You can use combination of ProbabilityDistribution and TruncatedDistribution as follows:

f[x_] := x/4
dist = ProbabilityDistribution[f[x], {x, 1, 3}];
td = TruncatedDistribution[{2, Infinity}, dist];

PDF[td, x]

$\begin{cases} \frac{2 x}{5} & 2<x<3 \\ 0 & \text{True} \end{cases}$

Probability[Conditioned[x <= t, x > 2], Distributed[x, dist]] // TeXForm

$\begin{cases} 1 & t\geq 3 \\ \frac{1}{5} \left(t^2-4\right) & 2<t<3 \end{cases}$

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Here's another method that uses the CDF to directly define the desired distribution and which might be a bit more intuitive/self-evident:

f[x_] := x/4
dist = ProbabilityDistribution[f[x], {x, 1, 3}];

ProbabilityDistribution[
 {
   "CDF", 
   Probability[x < y \[Conditioned] x > 2, x \[Distributed] dist]
 },
 {y, -\[Infinity], \[Infinity]}
];

ProbabilityDistribution[Piecewise[{{(2*\[FormalX])/5, 2 < \[FormalX] < 3}}, 0], {\[FormalX], -Infinity, Infinity}]

In other words: we just compute the CDF with Probability and then plonk it into ProbabilityDistribution while telling it that it's a CDF rather than a PDF. This is a useful trick to keep in mind, since the CDF is sometimes easier to work with since it's a proper probability.

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  • $\begingroup$ nice! (I felt that there must be a more direct way out there!) $\endgroup$ – Conor Cosnett Jul 2 at 14:16
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Sometimes "compact" can be considered elegant:

(x/4)/Integrate[x/4, {x, 2, 3}]
(* (2 x)/5 *)

If you need it in terms of Mathematica's ProbabilityDistribution:

d = ProbabilityDistribution[(x/4)/Integrate[x/4, {x, 2, 3}], {x, 2, 3}]
PDF[d, x]

Conditional density

CDF[d, x]

Conditional distribution function

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I learned from @kglr that Probability can give you a conditional CDF. One can take the derivative of this to compute the desired conditional PDF.

A = X >= 2;
conditionalCDF = Probability[X <= y \[Conditioned] A, X \[Distributed] ProbabilityDistribution[x/4, {x, 1, 3}]];
D[conditionalCDF, y]
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