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# Non 1-1 transformation of continuous random variable watch

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

So,

fX(x) = se-sx, x 0
0, otherwise

FX(x) = 1 - e-sx, x 0
0 otherwise

Let Y = sin X

FY (y) = P(Y y)
= P(sinX Y)
= P(X arcsin(y), X - arcsin(y)) {This is where I become slightly unsure}
=FX(arcsin(y)) - FX(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?
2. (Original post by 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.
3. (Original post by 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 /6 and 5/6 I get e-s/6 - e-s5/6

Similarly between 2 and 4 I get e-s13/6 - e-s17/6

So summing to infinity, I get e-s/6 - e-s5/6 + e-s13/6 - e-s17/6 + e-s25/6 - e-s29/6

So, it is e-s(1+12n)/6 - e-s(5+12n)/6, both from n=0 to infinity.

Calculating this on Maple I get 1/(es/6(1-(1/e2s))) - 1/(e5s/6(1-(1/e2s)))
4. 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, 0x/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 fX(x) = 2/ , 0x/2 and fX (x) = 0 otherwise.

So FX (x) = 0, x<0
= 2x/, 0x/2
=1, x>/2

Y=4SinX

FY(y)= P(Yy)
=P(4sinX y)
=P( - arcsin(y/4) X 2 - arcsin(y/4)) (*)
= FX (2 - arcsin(y/4)) - FX ( - arcsin(y/4)
= 2 + (2/)arcsin(y/4)

So fY(y) = (2/)(1/4)(1/) for 0y4 and 0 otherwise.

Is this ok so far? I'm very unsure of the stage marked by (*).
5. (Original post by Will.Honeyman)
=P( - arcsin(y/4) X 2 - arcsin(y/4)) (*)
= FX (2 - arcsin(y/4)) - FX ( - arcsin(y/4)
= 2 + (2/)arcsin(y/4)

So fY(y) = (2/)(1/4)(1/) for 0y4 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.
6. That makes a lot of sense, thanks. Carrying on I get E(Y) = 8/Pi which seems reasonable.
7. (Original post by Will.Honeyman)
That makes a lot of sense, thanks. Carrying on I get E(Y) = 8/Pi which seems reasonable.
For what it's worth, I agree with that.

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