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  1. #1
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    pdf

    Find the mean and variance for  X given the pdf  f(x) = .15e^{-0.15(x-0.5)}, \ x \geq .5

    If  Y = X-0.5 find the mean and variance of  Y .

    So  M_{X}(t) = \int_{0.5}^{\infty} .15e^{-0.15(x-0.5)}e^{tx} \ dx  = .15e^{0.75} \int_{0.5}^{\infty} e^{x(t-0.15)} \ dx

     M_{X}(t) = \frac{.15e^{0.5}}{t-0.15}e^{x(t-0.15)} evaluated from  x = 0.5 to  x = \infty .

    I have a feeling that this is to messy just to find the mean and variance.

    And to find the mean and variance for  Y, I wrote  X = Y+0.5 , and so  f_{Y}(y) = .15e^{-0.15(y-4.5)}, \ y \geq 0.5 . Then do the same thing.
    Last edited by heathrowjohnny; February 19th 2008 at 01:31 PM.
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    Find the mean and variance for  X given the pdf  f(x) = .15e^{-0.15(x-5)}, \ x \geq .5

    If  Y = X-0.5 find the mean and variance of  Y .

    So  M_{X}(t) = \int_{0.5}^{\infty} .15e^{-0.15(x-5)}e^{tx} \ dx  = .15e^{0.75} \int_{0.5}^{\infty} e^{x(t-0.15)} \ dx

     M_{X}(t) = \frac{.15e^{0.5}}{t-0.15}e^{x(t-0.15)} evaluated from  x = 0.5 to  x = \infty .

    I have a feeling that this is to messy just to find the mean and variance.

    And to find the mean and variance for  Y, I wrote  X = Y+0.5 , and so  f_{Y}(y) = .15e^{-0.15(y-4.5)}, \ y \geq 0.5 . Then do the same thing.
    Are you sure your density function isn't  f(x) = .15e^{-0.15(x-.5)}, \ x \geq .5 ? If it is, then you have an exponential distribution of the form f(x) = \frac{1}{\theta }e^{ - \frac{x}{\theta }}, where E(x) = \theta and Var(x) = \theta ^2. I found \theta but I am sure you can on your own.
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  3. #3
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    ahhh, my mistake, it was  0.5 . Then the mgf is already determined.

    Thanks
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  4. #4
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    Well, after the chairs problem I screwed up I am glad I can actually be of help to you.
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  5. #5
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    actually, is this an exponential distribution?  f(x) = 0.15e^{-0.15(x-0.5)} = 0.32e^{-0.15x} .
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  6. #6
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    0.15e^{ - 0.15(x - 0.5)}  = 0.15e^{ - 0.15x + 0.075}  = 0.15e^{ - 0.15x} e^{0.075}

    0.15e^{ - 0.15x} e^{0.075}  \approx 0.15e^{ - 0.15x} (1.077884) \approx 0.1616826e^{ - 0.15x} \ldots \ldots

    It is easier if you convert the decimals to fractions first, but you definitely did something wrong.


    \int\limits_{\frac{1}{2}}^\infty  {\frac{3}{{20}}} e^{ - \frac{3}{{20}}(x - \frac{1}{2})} dx = \int\limits_0^\infty  {\frac{3}{{20}}} e^{ - \frac{3}{{20}}(x)} dx = 1
    It is an exponential distribution shifted to the right by \frac{1}{2}.
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  7. #7
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    ahhh now I see, thanks a lot.
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  8. #8
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    and for  Y = X - 0.5 , then  X = Y+ 0.5 and  f_{Y}(y) = \frac{3}{20}e^{-\frac{3}{20}y}, \ y \geq \frac{1}{2} . But this is not an exponential distribution right? Because now we have  \int\limits_{\frac{1}{2}}^{\infty} \frac{3}{20}e^{-\frac{3}{20}y} \ dy = \int\limits_{0}^{\infty}  \frac{3}{20}e^{-\frac{3}{20}(y + \frac{1}{2})} \ dy
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  9. #9
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    Quote Originally Posted by heathrowjohnny View Post
    and for  Y = X - 0.5 , then  X = Y+ 0.5 and  f_{Y}(y) = \frac{3}{20}e^{-\frac{3}{20}y}, \ y \geq \frac{1}{2} . But this is not an exponential distribution right? Because now we have  \int\limits_{\frac{1}{2}}^{\infty} \frac{3}{20}e^{-\frac{3}{20}y} \ dy = \int\limits_{0}^{\infty}  \frac{3}{20}e^{-\frac{3}{20}(y + \frac{1}{2})} \ dy
    You don't need to worry about that (it looks incorrect BTW) if you know E(X) and Var(X). Just use these formulas:

    E(aX + b) = aE(X) + b and Var(aX + b) = a^2 Var(x)

    Remember that for f(x) = \frac{1}{\theta }e^{ - \frac{x}{\theta }}<br />
, E(x) = \theta and Var(x) = \theta ^2. So you have E(X)=\frac{20}{3} and Var(X)=\frac{400}{9}
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