Non 1-1 transformation of continuous random variable

X is exponentially distributed with mean s.
Find P(Sin(X)> 1/2)

So,

f_X(x) = se^-sx, x

\geq

0
0, otherwise

FX(x) = 1 - e^-sx, x

\geq

0
0 otherwise

Let Y = sin X

F_Y (y) = P(Y

\leq

y)
= P(sinX

\leq

Y)
= P(X

\leq

arcsin(y), X

\geq

\pi

- arcsin(y)) {This is where I become slightly unsure}
=F_X(arcsin(y)) - F_X(Pi - arcsin(y))
=1-e-^s(arcsin(y))- (1-e-^{s( - arcsin(y))})

From here I can differentiate to find the pdf and then use that to find sinX < 1/2. But I don't think this is right so far?

Reply 1

ghostwalker

17

Will.Honeyman

...

So, you're going to need to work out for what values of X, Sin(X) is > 1/2.

In the interval 0 to 2pi, it's ....

So integrate your pdf between those limits.

NOW, what about the interval 2pi to 4pi, etc.

Then sum to infinity.

Reply 2

Will.Honeyman

OP

ghostwalker

So, you're going to need to work out for what values of X, Sin(X) is > 1/2.

In the interval 0 to 2pi, it's ....

So integrate your pdf between those limits.

NOW, what about the interval 2pi to 4pi, etc.

Then sum to infinity.

Thanks.

When I integrate between

\pi

/6 and 5

\pi

/6 I get e^{-s $\pi$ /6} - e^{-s5 $\pi$ /6}

Similarly between 2

\pi

and 4

\pi

I get e^{-s13 $\pi$ /6} - e^{-s17 $\pi$ /6}

So summing to infinity, I get e^{-s $\pi$ /6} - e^{-s5 $\pi$ /6} + e^{-s13 $\pi$ /6} - e^{-s17 $\pi$ /6} + e^{-s25 $\pi$ /6} - e^{-s29 $\pi$ /6}

So, it is

\sum

e^{-s $\pi$ (1+12n)/6} -

\sum

e^{-s $\pi$ (5+12n)/6}, both from n=0 to infinity.

Calculating this on Maple I get 1/(e^{s $\pi$ /6}(1-(1/e^{2s $\pi$}))) - 1/(e^{5s $\pi$ /6}(1-(1/e^{2s $\pi$})))

Reply 3

Will.Honeyman

OP

I'm also working on a similar problem:

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)?

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/

\sqrt{1-(y^2)/4}

) for 0

\leq

y

\leq

4 and 0 otherwise.

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

Reply 4

ghostwalker

17

Will.Honeyman

=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/

\sqrt{1-y[sup]2[/sup]/4}

) for 0

\leq

y

\leq

4 and 0 otherwise.

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

I don't claim any great expertise on this, so bear that in mind.

My thinking would be for (*)

=P( 0 < X < arcsin(y/4))

bearing in mind the restrictions on X.