How to find the density function of a random variable with a given distribution?

[nhrock3]

$$Y-U(-2\pi,2\pi)$$
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 ??

 

[HallsofIvy]

Please, tell us what you are talking about!

I think that you are saying that uniformly distributed between $-2\pi$ and [itex]and $2\pi$, but I have to guess that bcause you didn't even say this was a probability question!<br /> <br /> <blockquote data-attributes="" data-quote="nhrock3" data-source="post: 2559729" cite="https://www.physicsforums.com/goto/post?id=2559729" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> nhrock3 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> $$Y-U(-2\pi,2\pi)$$<br /> find the density function of z=tan(Y)<br /> ? </div> </div> </blockquote> What have you done? You know that you are to show what efforts you have already made don't you?<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> i had a similar question<br /> <br /> X-U(0,1)<br /> find the density function of W=a+bx<br /> the solution is<br /> W-U(a,a+b)<br /> <br /> how to solve the first question ?? </div> </div> </blockquote> What is the density function for Y?[/itex]

 

[nhrock3]

it is probability question

the density function of Y is distributed evenly
$$Y-U(-2\pi,2\pi)$$

i tried to solve it like the example question i showed

but here in tangense i have no idea
because i could find the density by this
(tan(-2pi),tan(2p))
but this is wrong because if we have an interval mutiplication streches it
subtraction moves it to the left
but tangense
i have no idea

 

[HallsofIvy]

Since Y itself is uniformly distributed from $-2\pi$ to $2\pi$, its cumulative probability function is $x/(2\pi)$ an its density function is the constant $dY/dx= 1/(2\pi)$. The density function of Z= tan(Y) is the derivative of tan(Y): $d(tan(x/(2\pi))$.

 

[nhrock3]

you said facts but how you get to them?
the final solution is
$$f_z(t)\frac{1}{\pi(1+t^2)}$$
so its like you said

but i can't see a logical way like in the solved example i showed
?