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  1. #1
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    pdf

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

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

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

    $\displaystyle M_{X}(t) = \frac{.15e^{0.5}}{t-0.15}e^{x(t-0.15)} $ evaluated from $\displaystyle x = 0.5 $ to $\displaystyle 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 $\displaystyle Y, $ I wrote $\displaystyle X = Y+0.5 $, and so $\displaystyle f_{Y}(y) = .15e^{-0.15(y-4.5)}, \ y \geq 0.5 $. Then do the same thing.
    Last edited by heathrowjohnny; Feb 19th 2008 at 12:31 PM.
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    Find the mean and variance for $\displaystyle X $ given the pdf $\displaystyle f(x) = .15e^{-0.15(x-5)}, \ x \geq .5 $

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

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

    $\displaystyle M_{X}(t) = \frac{.15e^{0.5}}{t-0.15}e^{x(t-0.15)} $ evaluated from $\displaystyle x = 0.5 $ to $\displaystyle 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 $\displaystyle Y, $ I wrote $\displaystyle X = Y+0.5 $, and so $\displaystyle 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 $\displaystyle f(x) = .15e^{-0.15(x-.5)}, \ x \geq .5 $? If it is, then you have an exponential distribution of the form $\displaystyle f(x) = \frac{1}{\theta }e^{ - \frac{x}{\theta }}$, where $\displaystyle E(x) = \theta$ and $\displaystyle Var(x) = \theta ^2$. I found $\displaystyle \theta$ but I am sure you can on your own.
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  3. #3
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    ahhh, my mistake, it was $\displaystyle 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? $\displaystyle f(x) = 0.15e^{-0.15(x-0.5)} = 0.32e^{-0.15x} $.
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  6. #6
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    $\displaystyle 0.15e^{ - 0.15(x - 0.5)} = 0.15e^{ - 0.15x + 0.075} = 0.15e^{ - 0.15x} e^{0.075} $

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

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


    $\displaystyle \int\limits_{\frac{1}{2}}^\infty {\frac{3}{{20}}} e^{ - \frac{3}{{20}}(x - \frac{1}{2})} dx = \int\limits_0^\infty {\frac{3}{{20}}} e^{ - \frac{3}{{20}}(x)} dx = 1$
    It is an exponential distribution shifted to the right by $\displaystyle \frac{1}{2}$.
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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 $\displaystyle Y = X - 0.5 $, then $\displaystyle X = Y+ 0.5 $ and $\displaystyle 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 $\displaystyle \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 $\displaystyle Y = X - 0.5 $, then $\displaystyle X = Y+ 0.5 $ and $\displaystyle 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 $\displaystyle \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 $\displaystyle E(X)$ and $\displaystyle Var(X)$. Just use these formulas:

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

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