$\displaystyle Y\sim U(-2\pi,2\pi)$
find the density function of z=tan(Y)
?
i had a similar question
X~U(0,1)
find the density function of W=a+bx
the solution is
W~U(a,a+b)
how to solve the first question ??
$\displaystyle Y\sim U(-2\pi,2\pi)$
find the density function of z=tan(Y)
?
i had a similar question
X~U(0,1)
find the density function of W=a+bx
the solution is
W~U(a,a+b)
how to solve the first question ??
READ below, I think the orginal density is WRONG.
USE $\displaystyle f_Z(z)=f_Y(y)\left| {dy\over dz}\right|$ which is easy to derive.
It's the derivatives of the CDFs, it's just calc 1.
So, $\displaystyle f_Y(y)={1\over 4\pi}$ on $\displaystyle (-2\pi, 2\pi)$
and $\displaystyle y=\arctan z$ giving you $\displaystyle {dy\over dz}={1\over 1+z^2}$
Hence $\displaystyle f_Z(z)={1\over 4\pi(1+z^2)}$
I THINK there is a problem here, we should be getting a Cauchy.
I think the original density should be a uniform on $\displaystyle (-\pi/2, \pi/2)$ because tangent isn't defined at $\displaystyle \pi/2$
The second one is obvious.
$\displaystyle f_X(x)=1$ on 0<x<1.
If you let W=a+bX then W exists on (a,a+b), by letting X=0 and X=1.
Next, I will show that W is uniform on that interval.
USE $\displaystyle f_W(w)=f_X(x)\left| {dx\over dw}\right|$
With w=a+bx we have $\displaystyle {dw\over dx}=b$
So, USE $\displaystyle f_W(w)=(1)\left| {1\over b}\right|={1\over b}$ IF b>0.
which is a uniform density.