0

What is the density function of $Y=1-X^3$, if $X$ is a Cauchy random variable?

My approach: $$Pr(Y<y)=Pr(1-X^{3}<y)=Pr(X<(1-y)^{-3})=\int^{y}_{-\infty}\left(1-\frac{1}{\pi(1+t^{2})}\right)^{-3}\:dt$$ - is this ok?

And then the density function would be the derivative of the function above?

EDIT1 $$Pr(Y<y)=Pr(1-X^{3}<y)=Pr(X>(1-y)^{-3})=1-\int^{(1-y)^{-3}}_{-\infty}\left(\frac{1}{\pi(1+t^{2})}\right)\:dt=1-A$$

The CDF for Cauchy distribution, according to Wikipedia Cauchy Distribution - do I use it correctly?

$$A=\frac{1}{\pi}arctan((1-y)^{-3})+\frac{1}{2} - \frac{1}{\pi}arctan(-\infty)-\frac{1}{2}=\frac{1}{\pi}arctan((1-y)^{-3})-\frac{1}{\pi}\frac{-\pi}{2}=\frac{1}{\pi}arctan((1-y)^{-3})+\frac{1}{2}$$

Than the distribution function would be the derivative of 1-A ?

is this correct?

Ryan
  • 309

1 Answers1

3

Apart from the problem mentioned by @Avitus in a comment, let me mention that the identity $$ P(X\lt g(x))=\int_{-\infty}^xg(f_X(t))\mathrm dt, $$ where $f_X$ is the density of $X$, is quite wrong. Please use the correct $$ P(X\lt g(x))=\int_{-\infty}^{g(x)}f_X(t)\mathrm dt. $$ When, as in your case, the function $g$ is increasing and differentiable, this is also $$ P(X\lt g(x))=\int_{-\infty}^{x}g'(s)f_X(g(s))\mathrm ds. $$ Likewise, if $g$ is decreasing and differentiable and $Y=g^{-1}(X)$, then $$ P(Y\lt x)=P(X\gt g(x))=\int_{g(x)}^{+\infty}f_X(t)\mathrm dt=\int_{-\infty}^{x}(-g'(s))f_X(g(s))\mathrm ds. $$

Did
  • 279,727
  • Thank you Did, could you also write the answer to the problem using the data in the question besides the general solution?That would really hep me understand it. I would appreciate your effort a lot! Thank you – Ryan Jun 23 '13 at 10:35
  • What about showing instead how @Avitus' comment and my answer helped you make some steps towards solving the question? This way we would know what is still blocking you. – Did Jun 23 '13 at 11:09
  • 1
    @Ryan Did is right: try to correct the inequality and the integral. It would be nice if you could add in your questions the computations you can write, before getting blocked. – Avitus Jun 23 '13 at 11:15
  • I added an Edit. Can you please verify if i'm using a correct approach. Is the distribution function the derivative of what I have obtained as the value of 1-A? – Ryan Jun 23 '13 at 11:45
  • Let me suggest to compare it to the (new) formula at the end of my post. – Did Jun 23 '13 at 12:03
  • Thanks Did, I was thinking that since the primitive of fX is known, since fX is the Cauchy Distribution, it would be easier than to change the variable and compute a new integral. If you look at what I did in the EDIT, does it look correct to you? – Ryan Jun 23 '13 at 12:29
  • Why one would first integrate the density $f_X$ and then derive the result escapes me, to tell you the truth. Using the formula $f_Y(x)=-g'(x)f_X(g(x))$ reached in my post seems much simpler and much less error-prone, no? – Did Jun 23 '13 at 13:45