Let's suppose we have the following location scale exponential distribution:
$P(X>x)={ e }^{ -C(x-{ x }_{ 0 }) }$
We are looking to determine $E(X)$ and $V(X)$
There are several ways to do it, but I'm looking at the fastest way to solve it.
What I've done:
I'm trying the get the density:
$F(x)=1-P(X>x)=1-{ e }^{ -C(x-{ x }_{ 0 }) }$
So, by derivation, $f(x)=C{ e }^{ -C(x-{ x }_{ 0 }) }$
From there, can we just say the following things : ???
$f(x)={ e }^{ C({ x }_{ 0 }) }\times C{ e }^{ -Cx }$
and since $C{ e }^{ -Cx }$ is just an exponential with parameter $C$, therefore with expectation $1/C$.
So $E(X)=\frac { { e }^{ C{ x }_{ 0 } } }{ C } $ and $V(X)=\frac { { e }^{ C{ x }_{ 0 } } }{ { C }^{ 2 } } $
Is it correct to formulate things that way ?
Thanks for your feedback.
Added :
By following another approach (the Gamma approach), I'm finding something weird:
$E(X)=\int _{ 0 }^{ +\infty }{ Cx{ e }^{ -C(x-{ x }_{ 0 }) }{ 1 }_{ x>{ x }_{ 0 } } } dx=C{ e }^{ C{ x }_{ 0 } }\int _{ 0 }^{ +\infty }{ x{ e }^{ -Cx }{ 1 }_{ x>0 } } dx=C{ e }^{ C{ x }_{ 0 } }\frac { 1! }{ { C }^{ 2 } } \int _{ 0 }^{ +\infty }{ \frac { { C }^{ 2 } }{ 1! } x{ e }^{ -Cx }{ 1 }_{ x>0 } } dx=\frac { { e }^{ C{ x }_{ 0 } } }{ C } $
Why do I get this wrong result ?
Best Answer
You could just standardize the variable, calculate its moments, then transform back: e.g., let $Y = g(X) = C(X-x_0)$, then $X = g^{-1}(Y) = Y/C + x_0$, where $Y \sim \operatorname{Exponential}(1)$. So $$\operatorname{E}[Y^k] = \int_{y=0}^\infty y^k e^{-y} \, dy = k!, \quad k = 1, 2, \ldots,$$ and it follows that $$\operatorname{Var}[Y] = 2 - 1 = 1,$$ hence $$\operatorname{E}[X] = 1/C + x_0, \quad \operatorname{Var}[X] = 1/C^2.$$
If you want to do the calculation using indicator functions, then you need to be mindful of the relationship between the indicator function and the interval of integration: that is to say, if your density is $$f_X(x) = Ce^{-C(x-x_0)} \mathbb 1(x > x_0),$$ then $$\operatorname{E}[X^k] = \int_{x=-\infty}^\infty x^k f_X(x) \, dx = \int_{x=-\infty}^\infty C x^k e^{-C(x-x_0)} \mathbb 1(x > x_0) \, dx = C \int_{x=x_0}^\infty x^k e^{-C(x-x_0)} \, dx.$$ (And yes, my choice of notation is very slightly different.) The point is that using an indicator function allows you to formalize the support of the distribution, and carry that over into your calculations. Now to finish evaluating the integral, you would do a substitution...which is tantamount to the same transformation that we discussed above.