# Average value and probability density

1. Sep 5, 2014

### andre220

1. The problem statement, all variables and given/known data

Consider a particle oscillating according to $x(t) = a\cos(\omega t)$:

Find $\rho(x)$, the probability density to find particle at position $x$.

Compute $\langle x\rangle$, $\langle x^2\rangle$.

2. Relevant equations

So in general we know that $\textrm{Prob} = \int f_X (x)\,dx$

Also that $\langle x\rangle = \int x f(x)\,dx$, where $f(x)$ is a PDF and

$\langle x^2\rangle = \int x^2 f(x)\,dx$.

3. The attempt at a solution

Ok so, I think I am just seriously overlooking something here. However, that being said I cannot think of a way to construct $\rho(x)$. We do know that it should normalize to 1, but I can't seem to think of where to get started at.

Any help is appreciated, thank you.

2. Sep 5, 2014

### D_Tr

Hint: the probability of finding the particle on a given interval depends on how much time the particle spends being in that interval.

3. Sep 5, 2014

### andre220

Thank you D_Tr for the response. I have been thinking, and I know in quantum mech. we have
$$\rho(x) = \left|\psi(x,t)\right|^2$$, so analogously, could we say that this same principle holds and then solve for [itex]\rho(x)[/tex] by normalizing it to one?

4. Sep 5, 2014

### D_Tr

I do not think you need to draw an analogy between this problem and quantum mechanics. When you need to find a probability distribution, you just need to find what the probability density is proportional to. Then you are basically done and you just need to normalize. In this case it is inversely proportional to 1/v(x).

5. Sep 6, 2014

### andre220

Ok, I think I have it (or well am at least on my way). It is not entirely intuitive to me, but here is what I have after some thinking. $$dt = \frac{dx}{dx/dt} = \frac{dx}{v(x)}$$. And like you said it is proportional to the inverse of the velocity. Therefore the probability of finding this particle in some region x is the ratio of the time (dt, above) and the total time for one traversal in the oscillator, i.e.: $$\rho(x)dx =\frac{dt}{\tau/2}=\frac{2}{\tau}\frac{dx}{v(x)} = \frac{2}{\tau}\frac{1}{v(x)}.$$ And all that is left is to find the normalization. I believe that this is correct, does my logic make sense?
Thanks.

6. Sep 6, 2014

### Ray Vickson

This problem is not well-specified. Where does the "randomness" come in, which you will need in order to have a probability density for $x$? I can think of two possible sources of randomness:
(1) The parameter $\omega$ is actually a random number, not a fixed value.
(2) The time $t$ is a random number.

In both cases, in order to give a probability density for $x$ you need a probability density for the underlying parameter, so you need either a density $\rho_{\Theta}(\theta)$ for $\theta$ or $\rho_T(t)$ for $t$. Note that you CANNOT just assume these are "uniformly distributed", because there is no such thing as a uniform distribution over the whole real line; you could, however, assume that $\theta$ has a uniform distribution over a finite interval, say over $[0, M]$ or that $t$ has a uniform distribution over a finite interval, say $[0,N]$. In the latter case you would be essentially equating $\rho(x) \Delta x$ with the average fraction of time that the graph of $y = \cos(\omega t)$ lies between $y = x$ and $y = x + \Delta x$ when $t$ is restricted to the interval $[0,N]$. You could even take the limit as $N \to \infty$ of this fraction, and maybe that is what your problem proposer means by a "probability density". This is somewhat in line with the suggestion that D_Tr gave you.

7. Sep 7, 2014

### D_Tr

@ Ray Vickson: I think it is fair to assume here that we choose random values of t with a uniform distribution over a time interval of, say, T/2. But of course the problem statement would be more complete if it included this and in addition it would make the undertaker aware of the fact that we could, for example, not sample uniformly in time.

@ andre220: What you wrote is correct, but you shouldn't remove the dx from the last equality. Also, if there is something that I could explain to make it clearer to you I would be glad. Essentially, what you have done is that firstly you calculated ρ(t) = 1/(T/2). This is already normalized since if you integrate both sides from 0 to T/2 with respect to t you get 1 on the right. ρ(t) is constant because of the assumption we made that we choose t uniformly, and the uniform distribution pdf is constant.
But you want to find ρ(x) so you make a change of variable. You have calculated ρ(t)dt and you want to find ρ(x)dx, which you have already done.

8. Sep 7, 2014

### andre220

Ok thank you both for your help and input. It definitely seems to me that it is an odd question but I do understand it now.