1. ## 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.

2. Originally Posted by heathrowjohnny
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.

3. ahhh, my mistake, it was $0.5$. Then the mgf is already determined.

Thanks

4. Well, after the chairs problem I screwed up I am glad I can actually be of help to you.

5. actually, is this an exponential distribution? $f(x) = 0.15e^{-0.15(x-0.5)} = 0.32e^{-0.15x}$.

6. $0.15e^{ - 0.15(x - 0.5)} = 0.15e^{ - 0.15x + 0.075} = 0.15e^{ - 0.15x} e^{0.075}$

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

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 $\frac{1}{2}$.

7. ahhh now I see, thanks a lot.

8. 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 $\int\limits_{\frac{1}{2}}^{\infty} \frac{3}{20}e^{-\frac{3}{20}y} \ dy = \int\limits_{0}^{\infty} \frac{3}{20}e^{-\frac{3}{20}(y + \frac{1}{2})} \ dy$

9. Originally Posted by heathrowjohnny
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 }}
$
, $E(x) = \theta$ and $Var(x) = \theta ^2$. So you have $E(X)=\frac{20}{3}$ and $Var(X)=\frac{400}{9}$