Total movement of bacteria assuming a random distribution

[fatpotato]

Homework Statement

Find an expression for the total movement $s$ of a bacteria knowing that its movement follow a normal distribution.

Relevant Equations

Expected value $\bar{x} = \int_{-\infty}^{\infty}x\cdot p(x)dx$
Gaussian function $p(x) = \frac{1}{\sqrt{2\pi}} \exp{-\frac{x^2}{2}}$

Hello,

I have to find an expression for the total movement of a bacteria $s$, knowing that the bacteria is placed (centered) on a two side ruler at position $x=0$ (so a negative $x$ value means the bacteria has moved to the left of the ruler) and that the probability it moves to $x$ is given by the Gaussian function :
$$p(x) = \frac{1}{\sqrt{2\pi}} \exp{-\frac{x^2}{2}}$$
I have to express the total displacement $s$ using the fact that $\bar{x} = \int_{-\infty}^{\infty}x\cdot p(x) dx$ is the average expected displacement. Of course, since $p(x)$ is centered, $\bar{x}$ equals to zero, which makes sense both mathematically and physically, but now I have trouble finding a connection between this and a total displacement.

I thought about using $s = \int_{-a}^{a}x\cdot p(x) dx$ where $a$ would be the total displacement at point $a$, but obviously, since $x\cdot p(x)$ is odd, any integral of this form will yield zero, so I thought about taking only one side with $s = \int_{0}^{a}x\cdot p(x) dx$, yet I am not convinced.

Evaluating this new integral gives the following result :
$$s = \int_{0}^{a}x\cdot p(x) dx = \frac{1}{\sqrt{2\pi}} \big( 1 - \exp{-\frac{a^2}{2}} \big)$$
This means that the total displacement lies in the interval $[0; \frac{1}{\sqrt{2\pi}})$, but why would be the total displacement of $s$ anything other than $s$ itself? My technique would imply that if we want to look at the total displacement $s$ of a bacteria, we have to first look at "all the movements up to point $a$".

Does this make any sense?

Thanks in advance.

Edit : Relevant equations formatting

 

[anuttarasammyak]

Don't you need time t as parameter for dispersion, e.g.　
$$P(x,t) =\frac{N}{2\sqrt{\pi}Dt}exp(-\frac{x^2}{4Dt})$$?

As for limited range integration, may I interpret it that we would omit active bacteria that go beyond this border ?

 

[fatpotato]

Don't you need time t as parameter for dispersion, e.g.　
$$P(x,t) =\frac{N}{2\sqrt{\pi}Dt}exp(-\frac{x^2}{4Dt})$$?

In this exercise, time is not considered at all (not even mentionned!).

Out of curiosity, is your equation used to model dispersion? What are $N$ and $D$ ?

 

[anuttarasammyak]

Thanks for your curiosity. N is source or number of participating bacteria. D is dispersion coefficient expressing activeness of bacteria.

As you said <x>=0. Possible candidates of your interest are
$$<|x|>=\int_{-\infty}^{+\infty}|x|P(x) dx= 2\int_0^{+\infty} x P(x)dx$$
$$<x^2>=\int_{-\infty}^{+\infty} x^2 P(x)dx$$
$$\sqrt{<x^2>}=\sqrt {\int_{-\infty}^{+\infty} x^2 P(x)dx}$$

 

[pasmith]

Even if time is not explicitly mentioned in the problem, I would interpret it at meaning that the bacteria moves a distance $X \sim N(0,\sigma^2)$ in one time unit, so that its total distance moved after $n$ time units is the sum of $n$ independent identically dstributed normal random variables.

 

[fatpotato]

As for limited range integration, may I interpret it that we would omit active bacteria that go beyond this border ?

I have trouble finding meaning myself. I suppose this could be a valid interpretation, although I don't know what my instructor has in mind.

Possible candidates of your interest are

Alas, we are not supposed to know about these equations (which are second order moment and RMS value if I am not mistaken), only the expected value equation has to be used.

Even if time is not explicitly mentioned in the problem, I would interpret it at meaning that the bacteria moves a distance $X \sim N(0,\sigma^2)$ in one time unit, so that its total distance moved after $n$ time units is the sum of $n$ independent identically dstributed normal random variables.

Would this translate to the equation I have written in my first post?