Mean and var of an exponential distribution using Fourier transforms

[Master1022]

Homework Statement

Derive the mean and the variance of the exponential distribution using the Fourier transform

Relevant Equations

$FT[f(t)] = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt$

Hi,

I was just thinking about different ways to use the Fourier transform in other areas of mathematics. I am not sure whether this is the correct forum, but it is related to probability so I thought I ought to put it here.

Question: Is the following method an appropriate way to calculate the mean and variance of the exponential distribution? Have I overlooked any technicalities to get to the result?

I know that the standard proof isn't hard, but I was wondering whether there are faster ways (saves having to do the integration by parts)

Approach:
So we use the following principles for the Fourier transform:
$FT[f(t)] = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt = F(\omega)$
$FT[t^n f(t)] = j^n \frac{d^n F}{d\omega^n}$

Using the second relation, we can see that:
$$ E[f(t)] = \int_{-\infty}^{\infty} t f(t) dt = \int_{-\infty}^{\infty} tf(t) e^{-j(0) t} dt = j \frac{d F}{d\omega} |_{\omega = 0} $$
and
$$ E[f^2] = \int_{-\infty}^{\infty} t^2 f(t) dt = \int_{-\infty}^{\infty} tf(t) e^{-j(0) t} dt = j^2 \frac{d^2 F}{d\omega^2} |_{\omega = 0} $$

We have $f(t) = \lambda e^{-\lambda t} u(t)$ (0 for $t < 0$) and we have the standard result that $FT[e^{-at} u(t)] = \frac{1}{a+j\omega}$. Therefore,
$$ FT[ \lambda e^{-\lambda t} u(t)] = \frac{\lambda}{\lambda+j\omega} $$
and
$$ E[f] = FT[ \lambda t e^{-\lambda t} u(t)] = j \frac{d }{d\omega} \left( \frac{\lambda}{\lambda+j\omega} \right)|_{\omega = 0} = \frac{\lambda}{(\lambda + j \omega)^2}|_{\omega = 0} = \frac{1}{\lambda} $$
Now we can do
$$ E[f^2] = FT[ \lambda t^2 e^{-\lambda t} u(t)] = j^2 \frac{d^2 }{d\omega^2} \left( \frac{\lambda}{\lambda+j\omega} \right)|_{\omega = 0} = \frac{2 \lambda}{(\lambda + j \omega)^3}|_{\omega = 0} = \frac{2}{\lambda^2} $$

Finally we can use: $Var[f] = E[f^2] - (E[f])^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}$

which yields the expected result.

Have I overlooked any details along the way or is this a valid way to derive the required quantities?

Thanks.

 

[jambaugh]

That looks good. You can take a look at how the characteristic function and the moment generating function of a probability distribution relate to each other and to it's Fourier transform for more insight.

 

[Master1022]

That looks good. You can take a look at how the characteristic function and the moment generating function of a probability distribution relate to each other and to it's Fourier transform for more insight.

Dear @jambaugh , thank you very much for responding! I will definitely look into those areas.

For anyone else looking at this forum, please note there is a slight typo in the post when I was copy and pasting the latex. The result is correct. The integral for the $E[f^2]$ should read (I forgot the $t^2$ in the middle integral and instead left it as $t$ when I copied the code from the $E[f]$ integral):

$$ E[f^2] = \int_{-\infty}^{\infty} t^2 f(t) dt = \int_{-\infty}^{\infty} t^2 f(t) e^{-j(0) t} dt = j^2 \frac{d^2 F}{d\omega^2} |_{\omega = 0} $$