Probability distributions and other variables.

[Beer-monster]

Homework Statement

I'm just checking my thinking on my understanding of probability and probability distributions as it seems a little rusty.

So let say we have a set of boxes arranged in a grid. We throw balls, all of identical mass m, onto the grid at random.First, am I right in saying that the probability of a ball landing in a given box can be given as the area of a single box a divided by the total area of the grid of boxes A.

$$p = \frac{a}{A}$$Secondly, if I know that the mean number of balls I throw in a given time is $\lambda = Np$ the probability of a given box containing n balls is given by the Poisson distribution.

$$P(n)= \frac{\lambda^{n}}{n!}e^{-\lambda}$$

If I wanted to determine the probability of the total mass inside a given box, how would I go about it.

My feeling is that as the mass is the same for all of the balls, the distribution of masses is the same as the distribution of the numbers. Therefore, can I just convert it to a function of total mass M by substitution of $n=M/m$ ?

i.e. $$P(M) = \frac{\lambda^{M/m}}{(M/m)!}e^{-\lambda}$$

Is this correct? If so that factor term seems a little odd now, is there a way to neaten it.

 

[glebovg]

Perhaps, you should consider the Multinomial distribution.

P(r₁ balls in box 1, ..., r_n balls in box n) is given by the multinomial distribution,

P(r₁, ..., r_n) = [1/n^r] * [r!/(r₁!...r_n!), where r = r₁ + ... + r_n.

 

[glebovg]

If we have n boxes then the probability of a ball landing in a given box is 1/n.

 

[Beer-monster]

Sorry I should have been more specific.

In this analogy I'm considering the boxes to not be in contact with each other. I.e. there is space between the boxes so the balls could potentially miss.

Also, I'm considering the distribution of potential outcomes for a single box not the set of boxes, therefore I'm not sure the multinomial distribution would apply.

 

[HallsofIvy]

So you really just have two outcomes- hitting that one given box and not hitting it. Sounds like a binomial distribution to me- p= area of box over entire area, q= 1- p.

 

[Beer-monster]

q is the probability of not hitting a box correct?

It is a binomial distribution, however, if we consider a large number of balls and we only know the average number of balls to land in a box in a certain time we can describe the system by a Poisson distribution?

Any thoughts on the issue of the mass. The various outcomes for total mass of a box should be distributed the same as the distribution for number. I'm just not 100% sure what that means with the algebra.

 

[Ray Vickson]

Sorry I should have been more specific.

In this analogy I'm considering the boxes to not be in contact with each other. I.e. there is space between the boxes so the balls could potentially miss.

Also, I'm considering the distribution of potential outcomes for a single box not the set of boxes, therefore I'm not sure the multinomial distribution would apply.

So, you have (n+1) "places": box 1, box 2, ..., box n and "space". If B = total area (boxes + space), then the probability that a ball falls into box 1 is p = a/B (a = area of box 1). If you toss N balls, the number falling into box 1 is binomially distributed with mean m = Np and variance Np(1-p). Whether or not this is approximately a Poisson distribution depends on N and p; the usual argument is that with N large and p small, with m = Np moderate, the distribution is nearly Poisson, and becomes exactly Poisson in the limit that p --> 0 and N --> infinity in such a way that Np remains constant.

It is also true that if the number of balls tossed in a certain period of time is Poisson with mean M (and the tosses are independent) then the number falling in box 1 will be Poisson with mean p*M; this is now exact, not an approximation.

RGV

 

[Beer-monster]

Thanks for the clarification it was very helpful.

Though I'm still wondering about the mass issue.

To boil it down, basically I'm wondering how to modify a probability distribution from a distribution of x to the distribution of a function of x, of the form x multiplied by a constant in this case.

For example, in physics if we work out the distribution of momentum we can convert that into a distribution of energy using

$E = \frac{p^{2}}{2m}$

Or if we wanted to derive the equation for Doppler spectral broadening we'd start with the Maxwell-Boltzmann distribution for particle velocity then substitute the Doppler relation for velocity and wavelength.

However, I'm not sure how that works for the Poisson distribution (or the binomial for that matter).

Could anyone point me the right way, is it just a matter of subsitution?

 

[Ray Vickson]

Thanks for the clarification it was very helpful.

Though I'm still wondering about the mass issue.

To boil it down, basically I'm wondering how to modify a probability distribution from a distribution of x to the distribution of a function of x, of the form x multiplied by a constant in this case.

For example, in physics if we work out the distribution of momentum we can convert that into a distribution of energy using

$E = \frac{p^{2}}{2m}$

Or if we wanted to derive the equation for Doppler spectral broadening we'd start with the Maxwell-Boltzmann distribution for particle velocity then substitute the Doppler relation for velocity and wavelength.

However, I'm not sure how that works for the Poisson distribution (or the binomial for that matter).

Could anyone point me the right way, is it just a matter of subsitution?

There are standard formulas connecting the density function f(p) of a random variable P and the density g(y) of a function Y = h(P); see. eg.,
http://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables .

RGV

 

[Beer-monster]

Thanks for the link Ray. It was very helpful, though I also had to dig out a couple of texbooks but between them I think I got the idea.

Last thought on this: If I wanted to calculate how many boxes had say 3 balls in. I would just calculate the distribution function f(n) for n=3 and multiply by the total number of boxes, correct?

i.e.

no. boxes with three balls $$= Nf(n=3)$$