Mean and variance of a cosine function

[iocal]

Hi guys, as the title suggests I am a lizzle puzzled here.
The exercise asks me to prove that the variance of a cosine function is 1/2 and that also requires calculating the expected value.
I think I am getting the limits of integration wrong. What I am trying is [0,2π]. Any help is greatly appreciated, thanks!

 

[iocal]

$\displaystyle \displaystyle \mu =\dfrac{1}{2\pi} \int_0^{2\pi} \cos{\theta}\ d\theta$

$\displaystyle \displaystyle \mathrm{V}[\theta] = \dfrac{1}{2\pi} \int_0^{2\pi} \cos^2{\theta}\ d\theta - \mu^2$....that is, mean of square minus mean squared

You should be able to justify your result for $\displaystyle \mu$ by looking at a graph of the cosine.

I am not familiar with your method. I simply tried to do it like that ∫xcosx dx .Integrate by parts to get 0 which seems quite reasonable looking at the graph.

Then for the variance ∫x^2cosx dx , the mean being zero. But the problem is that I got the variance to be equal to 4π evaluating the integral over [0,2π]. Any thoughts?

 

[DrPhil]

I am not familiar with your method. I simply tried to do it like that ∫xcosx dx .Integrate by parts to get 0 which seems quite reasonable looking at the graph.

Then for the variance ∫x^2cosx dx , the mean being zero. But the problem is that I got the variance to be equal to 4π evaluating the integral over [0,2π]. Any thoughts?

My previous post was completely off the wall! Sorry about that - please ignore that post.

OK. The cosine distribution is limited from -pi/2 to +pi/2, because negative frequencies are meaningless. The normalization factor is
$\displaystyle \displaystyle \int_{-\pi/2}^{\pi/2} \cos{x}\ dx = \sin{x}\mid_{-\pi/2}^{\pi/2}\ = 2$

First moment of x = $\displaystyle \displaystyle \dfrac{1}{2}\int_{-\pi/2}^{\pi/2}x\ \cos{x}\ dx = 0$...because it is an odd function

Second moment = $\displaystyle \displaystyle \dfrac{1}{2}\int_{-\pi/2}^{\pi/2}x^2\ \cos{x}\ dx = \int_0^{\pi/2}x^2\ \cos{x}\ dx$

With the different limits, I seem to come up with $\displaystyle [(\pi/2)^2 - 2]$.

Makes me wonder how they define the cosine distribution. Could it be
$\displaystyle f(x) = \frac{\pi}{4}\ \cos\left( \frac{\pi}{2}x \right)$,...for $\displaystyle -1 \leq x \leq 1$ ?

 

[iocal]

My previous post was completely off the wall! Sorry about that - please ignore that post.

OK. The cosine distribution is limited from -pi/2 to +pi/2, because negative frequencies are meaningless. The normalization factor is
$\displaystyle \displaystyle \int_{-\pi/2}^{\pi/2} \cos{x}\ dx = \sin{x}\mid_{-\pi/2}^{\pi/2}\ = 2$

First moment of x = $\displaystyle \displaystyle \dfrac{1}{2}\int_{-\pi/2}^{\pi/2}x\ \cos{x}\ dx = 0$...because it is an odd function

Second moment = $\displaystyle \displaystyle \dfrac{1}{2}\int_{-\pi/2}^{\pi/2}x^2\ \cos{x}\ dx = \int_0^{\pi/2}x^2\ \cos{x}\ dx$

With the different limits, I seem to come up with $\displaystyle [(\pi/2)^2 - 2]$.

Makes me wonder how they define the cosine distribution. Could it be
$\displaystyle f(x) = \frac{\pi}{4}\ \cos\left( \frac{\pi}{2}x \right)$,...for $\displaystyle -1 \leq x \leq 1$ ?

Doesn't explain much but if it helps the context is the periodogram of a function. In any case thanks a lot.

 

[iocal]

But why do you use the scaling factor 1/2 in the integral? I have not seen that before.

 

[DrPhil]

But why do you use the scaling factor 1/2 in the integral? I have not seen that before.

The distribution function f(x) has to integrate to unity. Since the integral of cosx is 2, the normalization factor is 1/2.

 

[iocal]

The distribution function f(x) has to integrate to unity. Since the integral of cosx is 2, the normalization factor is 1/2.

Got you. By the way if you have a good advanced calculus book to suggest please pm me.