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

    So,

    fX(x) = se-sx, x \geq0
    0, otherwise

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

    Let Y = sin X

    FY (y) = P(Y \leqy)
    = P(sinX \leq Y)
    = P(X \leq arcsin(y), X \geq \pi- 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?
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    (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.
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    (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 \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/(es\pi/6(1-(1/e2s\pi))) - 1/(e5s\pi/6(1-(1/e2s\pi)))
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    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\leqx\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 fX(x) = 2/\pi , 0\leqx\leq\pi/2 and fX (x) = 0 otherwise.

    So FX (x) = 0, x<0
    = 2x/\pi, 0\leqx\leq\pi/2
    =1, x>\pi/2

    Y=4SinX

    FY(y)= P(Y\leqy)
    =P(4sinX \leqy)
    =P(\pi - arcsin(y/4) \leq X \leq 2\pi - arcsin(y/4)) (*)
    = FX (2\pi - arcsin(y/4)) - FX (\pi - arcsin(y/4)
    = 2 + (2/\pi)arcsin(y/4)

    So fY(y) = (2/\pi)(1/4)(1/\sqrt{1-(y^2)/4}) for 0\leqy\leq4 and 0 otherwise.

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

    So fY(y) = (2/\pi)(1/4)(1/\sqrt{1-y<sup>2</sup>/4}) for 0\leqy\leq4 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.
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    That makes a lot of sense, thanks. Carrying on I get E(Y) = 8/Pi which seems reasonable.
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    (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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