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Math Help - Uniform/ exponential dist.

  1. #1
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    Uniform/ exponential dist.

    I think these are the last two I can't get, from the four sections. I don't really even know where to start:

    1.) Let X have an exponential dist. with mean, theta > 0. Show that

    P(X > x + y | X > x) = P(X > y)

    --------

    2.) Let Y have a uniform dist. U (0, 1)

    and let W=a +(b-a)Y a<b


    a) Find the dist. function of W. [Hint: find P[a+ (b-a)Y < or =w]

    b) How is W distributed?
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  2. #2
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    1.) Let X have an exponential dist. with mean, theta > 0. Show that

    P(X > x + y | X > x) = P(X > y)
    remember that
     P(X > x + y | X > x) = \frac{P(X>x+y)}{P(X>x)}

    P(X>a) =1-F(a) where F is the cdf of X.
    This is where to start. By the sound of it I think you will be able to continue nicely.

    2.) Let Y have a uniform dist. U (0, 1)

    and let W=a +(b-a)Y a<b


    a) Find the dist. function of W. [Hint: find P[a+ (b-a)Y < or =w]

    b) How is W distributed?
    Start by finding the distribution function of Y. Then note that the function W(Y) is strictly increasing. Can you continue?
    Last edited by badgerigar; July 5th 2008 at 05:38 PM. Reason: fixed significant typo; had pdf instead of cdf
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  3. #3
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    Nope, I still don't understand. Sorry. I'm not sure what to do next.
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  4. #4
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    Quote Originally Posted by amor_vincit_omnia View Post
    I think these are the last two I can't get, from the four sections. I don't really even know where to start:

    1.) Let X have an exponential dist. with mean, theta > 0. Show that

    P(X > x + y | X > x) = P(X > y)

    --------

    2.) Let Y have a uniform dist. U (0, 1)

    and let W=a +(b-a)Y a<b


    a) Find the dist. function of W. [Hint: find P[a+ (b-a)Y < or =w]

    b) How is W distributed?
    1.) f(x) = \frac{1}{\theta} \, e^{-\frac{x}{\theta}}, ~ x \geq 0, and zero otherwise.

    Following from what badgerigar said:

    \Pr(X > x + y | X > x) = \frac{ 1 - \int_{0}^{x+y} \frac{1}{\theta} \, e^{-\frac{t}{\theta}} \, dt }{1 - \int_{0}^{x} \frac{1}{\theta} \, e^{-\frac{t}{\theta}} \, dt } = \frac{ \theta - \int_{0}^{x+y}e^{-\frac{t}{\theta}} \, dt }{\theta - \int_{0}^{x}e^{-\frac{t}{\theta}} \, dt } = ......


    \Pr(X > y) = 1 - \int_{0}^{y} \frac{1}{\theta} \, e^{-\frac{t}{\theta}} \, dt = .....


    You should get the same answer for both.

    --------------------------------------------------------------------------

    2.) Many aproaches are possible. Here's one:

    Consider the cdf of W: F(w) = \Pr(W < w) = \Pr(a + (b-a)Y < w) = \Pr \left( Y < \frac{w - a}{b - a} \right) = \int_{0} ^{\frac{w-a}{b-a}} 1 \, dy = ..... , a \leq w \leq b (since 0 \leq y \leq 1 ) and zero otherwise.

    Then f(w) = \frac{dF}{dw} = ....... .

    Unsurprisingly, it's seen that W ~ U(a, b) ......
    Last edited by mr fantastic; July 5th 2008 at 07:01 PM. Reason: Added a bit more to Q1 and replaced a W with w
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