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Math Help - Sum of two uniform distributions and other questions.

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    Sum of two uniform distributions and other questions.

    Assume that
    X has uniform distribution on the interval [0,1] and Y has uniform
    distribution on the interval [0,2]. Find the density of Z = X + Y .


    Would this just be 1+1/2 = 3/2?


    Suppose X with values (0, 1) has density \[f(x) = cx^2(1-x)^2\] for 0 < x < 1. Find:

    a) the constant c; b) E[X]; c) Var[X]

    I get, a) c = 30; b) 1/2; c) 1/28

    Transistors produced by one machine have a lifetime which is exponentially distributed with mean 100 hours. Those produced by a second machine have an exponentially distributed lifetime with mean 200 hours. A package of 12 transistors contains 4 produced by the first machine, and 8 produced by the second machine. Let X be the lifetime of a transistor picked at random from this package. Find:

    a) P(X greater or equal to 200 hours); b) E[X]; c) Var[X]

    So would E[X] = (1/3)(100) + (2/3)(200) = 500/3? If so, then for a) I get e^(-6/5) = 0.3012, and for c) I get (500/3)^2 = 250000/9

    Last edited by mr fantastic; November 27th 2010 at 09:32 PM. Reason: Re-titled.
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  2. #2
    MHF Contributor harish21's Avatar
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    Thats too many questions in a single thread. Try to post only one question in a thread. Its easy to read and makes your thread less messy.

    1) I dont think so. Look at this link(page No 8) that finds the pdf of sum of 2 independent uniform RVs.

    2)the values look correct.

    Edit: No 3.. Your expected value is correct.. not the variance.. You are doing V(X)=(E(X))^2, which is not correct.

    You have found E[X]..then you need to find E[X^2] (using conditional expectation) and the variance is found the following identity:

    V(X)=E(X^2)-E^2(X)
    Last edited by harish21; November 27th 2010 at 01:48 PM.
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    But for exponential r.v.'s isn't E[X]=SD[X]? So then to get variance, couldn't you just square SD[X]?
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    MHF Contributor matheagle's Avatar
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    the sum of two indep uniforms is triangular
    the easiest technique to obtain the CDF of the sum via geometry on the unit square
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    I got the sum of the RVs, thanks!

    Could someone verify the third question? Is the logic correct in my previous post?
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    MHF Contributor matheagle's Avatar
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    what is the third question?
    The density x^2(1-x)^2 on 0<x<1 is a Beta, so all those questions are obvious.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by BrownianMan View Post
    Assume that
    [LEFT]X has uniform distribution on the interval [0,1] and Y has uniform
    distribution on the interval [0,2]. Find the density of Z = X + Y .


    Would this just be 1+1/2 = 3/2?
    In general if \phi_{x} (*) is the PDF of a random variable x and \phi_{y} (*) is the PDF of a random variable y , the PDF od the random variable z=x+y is given by...

    \displaystyle \phi_{z} (z) = \phi_{x} (z) * \phi_{y} (z) (1)

    ... where * means 'convolution'. The (1) permits to use fuitfully L-transform od F- transform tecniques to obtain in confortable way the \phi_{z} (*)...

    Kind regards

    \chi \sigma
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    In general if \phi_{x} (*) is the PDF of a random variable x and \phi_{y} (*) is the PDF of a random variable y , the PDF od the random variable z=x+y is given by...

    \displaystyle \phi_{z} (z) = \phi_{x} (z) * \phi_{y} (z) (1)

    ... where * means 'convolution'. The (1) permits to use fuitfully L-transform od F- transform tecniques to obtain in confortable way the \phi_{z} (*)...
    In the problem proposed by BrownianMan is...

    \displaystyle \mathcal{L} \{\phi_{x} (*)\} = \Phi_{x} (s) = \frac{1-e^{-s}}{s}

    \displaystyle \mathcal{L} \{\phi_{y} (*)\} = \Phi_{y} (s) = \frac{1-e^{-2 s}}{2 s} (1)

    ... so that is...

    \displaystyle \phi_{z} (z)= \mathcal{L}^{-1}\{\frac{1-e^{-s} - e^{-2 s} + e^{-3 s}}{2 s^{2}} \} (2)

    The function \phi_{z} (z) is illustrated here...



    Kind regards

    \chi \sigma
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    this is the third question:

    Transistors produced by one machine have a lifetime which is exponentially distributed with mean 100 hours. Those produced by a second machine have an exponentially distributed lifetime with mean 200 hours. A package of 12 transistors contains 4 produced by the first machine, and 8 produced by the second machine. Let X be the lifetime of a transistor picked at random from this package. Find:

    a) P(X greater or equal to 200 hours); b) E[X]; c) Var[X]

    So would E[X] = (1/3)(100) + (2/3)(200) = 500/3? If so, then for a) I get e^(-6/5) = 0.3012, and for c) I get (500/3)^2 = 250000/9



    It is independent of the others.
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  10. #10
    MHF Contributor harish21's Avatar
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    As told in post No 2, the expected value is correct, but the variance is wrong.
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