0
$\begingroup$

Creating the following transformed distribution $A$

A = TransformedDistribution[Sqrt[x^2], x \[Distributed] NormalDistribution[\[Mu], \[Sigma]]]

and then using A to create another TransformedDistribution $B$,

B = TransformedDistribution[1/2 + Sqrt[c^2 d^2 (d^2 - c x^2)]/(2 c d^2), x \[Distributed] A]

seems to get Mathematica stuck on

PDF[B, x]

Or is it me doing something wrong?

$\endgroup$
  • $\begingroup$ In general, finding the PDF is likely to involve integration. If Mathematica struggles with the integral, it will struggle with the PDF. It may return useful results for numerical values of x $\endgroup$ – mikado Oct 7 '18 at 13:03
  • $\begingroup$ Could I generate many values for $A$, and produce the corresponding values for $B$ using Mathematica? If so, how? $\endgroup$ – user120911 Oct 7 '18 at 13:27
3
$\begingroup$

In the general case the random variable B is complex-valued. It should also be noticed that B depends on four parameters. All that is too hard even for the human mind. Mathematica answers the question for concrete values of c and d, eg

A = TransformedDistribution[RealAbs[x],x \[Distributed] NormalDistribution[\[Mu],[Sigma]]]
B = TransformedDistribution[1/2 + Re[Sqrt[ (1 - x^2)]/(2 )],x \[Distributed] A];
PDF[B,x]//TeXForm

$$\begin{array}{cc} \{ & \begin{array}{cc} \frac{(2 x-1) e^{-\frac{\mu ^2-4 x^2+4 x}{\sigma ^2}} \left(e^{\frac{\left(\mu -2 \sqrt{-(x-1) x}\right)^2}{2 \sigma ^2}}+e^{\frac{\left(\mu +2 \sqrt{-(x-1) x}\right)^2}{2 \sigma ^2}}\right)}{\sqrt{2 \pi } \sigma \sqrt{-(x-1) x}} & \frac{1}{2}<x<1 \\ 0 & x>1\lor x<\frac{1}{2} \\ \text{Indeterminate} & \text{True} \\ \end{array} \\ \end{array} $$

The result of

PDF[TransformedDistribution[Im[Sqrt[ (1 - x^2)]/(2 )],x \[Distributed] A], x]

is similar.

$\endgroup$
3
$\begingroup$

If you restrict the values of $c$ and $d$ to real values where $c<0$ and $d\neq 0$, then all values of the desired random variable are real and a symbolic solution exists. (With more work maybe other symbolic solutions exist if the conditions on $c$ and $d$ are relaxed - I won't tackle that.)

We see that with $c<0$ and $d\neq 0$

FullSimplify[1/2 + Sqrt[c^2 d^2 (d^2 - c x^2)]/(2 c d^2), Assumptions -> {c < 0, d != 0}]

$$\frac{1}{2}-\frac{\sqrt{d^4-c d^2 x^2}}{2 d^2}$$

Because of the restrictions we can simplify this to

$$\frac{1}{2}-\frac{1}{2} \sqrt{1-\frac{c x^2}{d^2}}$$

Because we know that $-c/d^2$ is positive, we can further simplify to

1/2 - Sqrt[1 + e x^2]/2

Now a symbolic solution can be obtained:

a = TransformedDistribution[Sqrt[x^2], x \[Distributed] NormalDistribution[μ, σ]];

b = TransformedDistribution[1/2 - Sqrt[1 + e x^2]/2, x \[Distributed] a,
    Assumptions -> {e > 0}];
PDF[b, z]

$$\begin{array}{cc} \{ & \begin{array}{cc} \frac{\frac{(1-2 z) e^{-\frac{\left(2 \sqrt{e} \sqrt{(z-1) z}+e\right)^2}{8 e^2}}}{\sqrt{e} \sqrt{(z-1) z}}+\frac{(1-2 z) e^{-\frac{1}{8} \left(1-\frac{2 \sqrt{(z-1) z}}{\sqrt{e}}\right)^2}}{\sqrt{e} \sqrt{(z-1) z}}}{2 \sqrt{2 \pi }} & z\leq 0 \\ 0 & \text{True} \\ \end{array} \\ \end{array}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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