Hello,
How to calculate this in Mathematica?
Suppose the probability density of random variable X is f(x), −∞<x<∞. what is the probability density of Y=X^3?
Sadly, it seems that Mathematica cannot handle generic symbolic distributions:
Probability[X > 0,
Distributed[X,
ProbabilityDistribution[f[x],
{x, -Infinity, Infinity}]]]
For an explicit distribution it works:
Probability[X^3 < t,
Distributed[X,
ProbabilityDistribution[
Sqrt[2]/Pi/(1 + x^4),
{x, -Infinity, Infinity}]]]
D[%, t] // Simplify
Okay. Thanks.
Did you see your results, in which there are many #1. Is it right?
(1/(4 \[Pi]))(2 \[Pi]-2 ArcTan[1-Sqrt[2] Root[-t+#1^3&,1]]+2 ArcTan[1+Sqrt[2] Root[-t+#1^3&,1]]-Log[1-Sqrt[2] Root[-t+#1^3&,1]+Root[-t+#1^3&,1]^2]+Log[1+Sqrt[2] Root[-t+#1^3&,1]+Root[-t+#1^3&,1]^2])
Another question:
D[Probability[Sin[X] < y,
Distributed[X, ProbabilityDistribution[2 x/(Pi^2), {x, 0, Pi}]]],
y] // Simplify
The output is
Why are there Inderterminate here and what is the meaning of that? I think this one should be omitted.
Those #1 are used to express algebraic numbers. You can check the documentation on Root. Actually, in that particular expression we find repeatedly Root[-t + #1^3 &, 1], which is simply the real cubic root of t. You can replace it with Surd[t, 3]:
FullSimplify[
Root[-t + #1^3 &, 1] == Surd[t, 3],
Element[t, Reals]]
The Indeterminate appears when y == 0 and when y == 1, where the derivative does not exist. You can plot the density function and see that it is discontinuous in those points:
D[Probability[Sin[X] < y,
Distributed[X,
ProbabilityDistribution[2 x/(Pi^2),
{x, 0, Pi}]]], y] // Simplify
Plot[%, {y, -1, 2}]
Thanks a lot.
- How to use
Piecewisefunction inProbabilityDistributionfunction? - Why is the
Indeterminaterange isTrueand what is the meaning ofTruehere?
Regards,
You could answer yourself by reading the documentation or experimentation. Here is an example of Piecewise probability density:
ProbabilityDistribution[
Piecewise[{{1 + x, -1 < x < 1},
{1/x, 1 < x < 2}}],
{x, -Infinity, Infinity},
Method -> "Normalize"]
The conditions in Piecewise are to be tested in the given order: the first that gives True is applied. The last condition is always True, and it applies when all the previous tests give False. It is the complement of the other tests. It can be confusing at first.
Thanks.
Do feel it is a little weird as in
ProbabilityDistribution[
Piecewise[{{1 + x, -1 < x < 1},
{1/x, 1 < x < 2}}],
{x, -Infinity, Infinity},
Method -> "Normalize"]
Outside of the Piecewise, there is {x, -Infinity, Infinity}. As Piecewise have given the range of the function, why should we add {x, -Infinity, Infinity} after it?
Another question:
D[Probability[Sin[X] < y,
Distributed[X,
ProbabilityDistribution[
Piecewise[{{2*x/(Pi^2), 0 < x < Pi}, {0,
x <= 0 || x >= Pi}}], {x, -Infinity, Infinity}]]],
y] // Simplify
outputs
Do you think this is the right answer? I think in this Piecewies limits, the answer should be
2/(\[Pi] Sqrt[1 - y^2]), 0<y<1 and 0, y equals others.
The {x, -Infinity, Infinity} looks redundant but it is required by the syntax of ProbabilityDistribution.
The Piecewise formula is perfectly correct, because it is the answer to the derivative of this function:
Probability[Sin[X] < y,
Distributed[X, ProbabilityDistribution[2 x/(Pi^2), {x, 0, Pi}]]]
Plot[%, {y, -1, 2}]
You can see from the plot that for y == 0 and for y == 1 the function is not differentiable. The derivative from the left and from the right are different.
Thanks.
I just feel that it is better to make ProbabilityDistribution function to use the range of Piecewise. For the second quesiton, I have a different view from you. From the original Probability Distribution fuction
ProbabilityDistribution[
Piecewise[{{2*x/(Pi^2), 0 < x < Pi}, {0,
x <= 0 || x >= Pi}}], {x, -Infinity, Infinity}]
which has stated that when x=0 or x>=Pi, the Probability Distribution is 0. So, when y=0 and y=1, its Probability Distribution should be 0 too.
May you check this:
The first
D[Probability[Sin[X] < y,
Distributed[X,
ProbabilityDistribution[
Piecewise[{{2*x/(Pi^2), 0 < x < Pi}, {0,
x <= 0 || x >= Pi}}], {x, -Infinity, Infinity}]]],
y] // Simplify
The second
D[Probability[Sin[X] < y,
Distributed[X,
ProbabilityDistribution[
Piecewise[{{2*x/(Pi^2), 0 < x < Pi}}], {x, -Infinity,
Infinity}]]], y] // Simplify
They have the same results. Not weird?
Hello,
Why doesn’t this one work:
PDF[TransformedDistribution[Sin[X],
X \[Distributed] ProbabilityDistribution[2*x/(Pi^2), 0 < x < Pi]],
y]
This one produces the
PDF[TransformedDistribution[Sin[X],
X \[Distributed] ProbabilityDistribution[2*x/(Pi^2), 0 < x < Pi]],
y]
Beware the syntax:
Probability[Sin[X] < t,
X \[Distributed] ProbabilityDistribution[2*x/(Pi^2), {x, 0, Pi}]]
D[%, t]
The density of a probability distribution is an object that is meant to be integrated on intervals. Its value in a single point does not matter at all for the distribution. In advanced calculus we call it a function defined almost everywhere.
The default value of Piecewise is zero. Hence the two functions
Piecewise[{{2*x/(Pi^2), 0 < x < Pi}, {0, x <= 0 || x >= Pi}}]
Piecewise[{{2*x/(Pi^2), 0 < x < Pi}}]
are the same thing.
Thanks.
But this one doesn’t work either.
PDF[TransformedDistribution[Sin[X],
X \[Distributed]
ProbabilityDistribution[2*x/(Pi^2), {x, 0, \[Pi]}]], y]
As it produces
PDF[TransformedDistribution[Sin[X],
X \[Distributed]
ProbabilityDistribution[(
2 \[FormalX])/\[Pi]^2, {\[FormalX], 0, \[Pi]}]], y]
Thanks.
Piecewiseuses default value val if none of theSubscript[cond, i]apply. The default for val is 0.
As the Subscript[cond, i] applies, such as 0 < x < Pi, so the default for val is not used here. They should not be the same.
Sorry, I don’t understand your question. “As the Subscript[cond, i] applies, such as 0 < x < Pi”: the condition applies to what?
Yes, it is sad that this does not work
PDF[TransformedDistribution[Sin[X],
X \[Distributed]
ProbabilityDistribution[2*x/(Pi^2), {x, 0, \[Pi]}]], y]
Hello,
Thanks for your kind reply and sorry for my delay reply. I have checked the document again. Now, I kne w the default 0 value for val. Before I had misunderstood it. So, there is no quesion now.
Best regards,
Okay. Thanks for your kind comfirmation.
Regards