Continuous random variable - transformation using sin

[Kate2010]

Homework Statement

There is a pin of length 4 which appear on a photograph, and the length of the image observed is y, an observation on the random variable Y. The pin is at an angle x, 0$$\leq$$x$$\leq$$$$\pi$$/2, to the normal to the film, this is an observation on the r.v. X.

1. If all angles X are equally likely then derive the distribution of Y.

2. What is E(Y)?

Homework Equations

The Attempt at a Solution

X is uniform so f_X(x) = 2/$$\pi$$ , 0$$\leq$$x$$\leq$$$$\pi$$/2 and f_X (x) = 0 otherwise.

So F_X (x) = 0, x<0
= 2x/$$\pi$$, 0$$\leq$$x$$\leq$$$$\pi$$/2
=1, x>$$\pi$$/2

Y=4SinX

F_Y(y)= P(Y$$\leq$$y)
=P(4sinX $$\leq$$y)
=P($$\pi$$ - arcsin(y/4) $$\leq$$ X $$\leq$$ 2$$\pi$$ - arcsin(y/4)) (*)
= F_X (2$$\pi$$ - arcsin(y/4)) - F_X ($$\pi$$ - arcsin(y/4)
= 2 + (2/$$\pi$$)arcsin(y/4)

So f_Y(y) = (2/$$\pi$$)(1/4)(1/[tex]\sqrt{1-y² /4}[/tex]) for 0$$\leq$$y$$\leq$$4 and 0 otherwise.

Is this ok so far? I'm very unsure of the stage marked by (*).

 

[lippyka]

E(Y)=\int_{0}^{4}yfY(y)dy=\int_{0}^{4}y(2/\pi)(1/4)(1/\sqrt{1-y2 /4}) dyI'm not sure how to do this integral, so any help is greatly appreciated!