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    Probability Distribution Function




    Hi Every one. Can any one help me to start solving this problem?i am stocked in the beginning.


    The range of a ballistic projectile is given by

    r=(v^2)sin2x/g

    where v is the initial velocity, x is the launch angle (relative to the horizontal) and g is the
    gravitational acceleration. The launch angle cannot be controlled precisely so is treated as a random variable x with f x(x). It should be clear that the maximum range occurs for x=0.25pi
    .

    #Determine an expression for the PDF of the range (now a random variable,r~f (r))
    assuming that x can never exceed the interval 0<x<0.5pi(0 and 0.5pi included)



    thanks
    Last edited by 22nelson; 03-14-2012 at 08:02 PM.

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    Re: Probability Distribution Function

    Do you mean now X is a random variable with bounded support (0, 0.5\pi), and e.g. X is a continuous random variable with a certain pdf, which you would like to know the pdf of R ?

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    Re: Probability Distribution Function

    actually i just found out i need to know the pdf os R which is equal to pdf of x in some point. i hope i did not make any one confuse
    thanks for your concern

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    Re: Probability Distribution Function

    I think you were saying that you figured it out? I don't know... I'm just confused...
    I don't have emotions and sometimes that makes me very sad.

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    Re: Probability Distribution Function

    no i did not

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    Re: Probability Distribution Function

    Assume you want to obtain the pdf of R in terms of the pdf of X.

    Then just use the Jacobian transform will do.

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    Re: Probability Distribution Function


    so do you think if i start like: P(R<a)
    it gives me the probability of R in point (a) which is equal to P(x). then derivative of that to get pdf
    am i in right track?

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