I have a random variable that gives me the number of hydrogen molecules in a spherical region. The random variable has a poisson pmf.

Now I have to define a new Random variable as the distance from the origin to the nearest molecule and find its pdf.

I am unable to think how to do this. I am unable to think of a way to relate the two random variables i.e. distance and number of molecules in the sphere?

Kindly guide me how to approach this problem. Thanks in advace.

  • $\begingroup$ So it seems to me that you essentially have a stochastic process: you have random variables $X_r$ for $r \geq 0$, such that $X_r$ is the number of molecules in the sphere of radius $r$ centered at the origin. Each $X_r$ is Poisson distributed with mean $f(r)$ where $f$ is some increasing function. (For instance, if there is a uniform overall density, then $f(r)=Cr^3$.) The $X_r$ are not independent; in particular if $R \geq r$ then $X_R \geq X_r$. And at the end you want to find $\inf \{ r \geq 0 : X_r \geq 1 \}$. Is this description correct? Can you give your $f$? $\endgroup$ – Ian Sep 29 '15 at 23:28
  • $\begingroup$ @Ian The sphere is not centered at the origin. It is anywhere in the space. We choose an arbitrary point in space as the center of our coordinate system. Now we are interested in a random variable X that defines the distance from the origin to the nearest molecule. All I have is the Poisson pmf of the random variable describing the number of molecules in the sphere and now I want to find the pdf of X. $\endgroup$ – Ashu Sanan Sep 29 '15 at 23:42
  • $\begingroup$ It being centered anywhere introduces some calculational technicalities, but is no major obstacle. My question was: you have the Poisson variable, but how does the mean of this variable depend on the radius of the sphere? The simplest thing that would make physical sense would be $f(r)=Cr^3$ for a constant $C$, so that you would have a uniform average density, but I could easily see it being something different depending on the context. $\endgroup$ – Ian Sep 29 '15 at 23:58
  • $\begingroup$ @Ian The distribution of the variable is directly dependent on the volume of the sphere: V. So I am assuming it to be uniform. $\endgroup$ – Ashu Sanan Sep 30 '15 at 0:14

Let the number in a ball of radius $R$ have Poisson distribution with parameter $\lambda$.

Let $X$ be the distance from the origin, or any point $P$, to the nearest molecule. Then $X\gt x$ if and only if the ball of radius $x$ with centre $P$ has no molecules. The number of molecules in a ball of radius $x$ has Poisson distribution parameter $\frac{\lambda x^3}{R^3}$. The probability this is $0$ is $\exp(-\lambda x^3/R^3)$.

So the cumulative distribution function of $X$ is $1-\exp(-\lambda x^3/R^3)$ (for $x\gt 0$). For the density function of $X$, we differentiate.

  • 1
    $\begingroup$ I think it should be made explicit that this is assuming that the average density is uniform (so that the ratio of the average number of molecules between a ball of radius $r$ and a ball of radius $R$ is the ratio of the volumes, i.e. $r^3/R^3$). This also assumes (apparently incorrectly) that the "center of the distribution" is the point from which we are measuring the distances. $\endgroup$ – Ian Sep 30 '15 at 0:00
  • $\begingroup$ There is no centre to the distribution. $\endgroup$ – André Nicolas Sep 30 '15 at 0:26
  • $\begingroup$ I do not mean the literal center, I mean the point where the spheres whose radius appear in the PMF are centered. Hence the scare quotes. $\endgroup$ – Ian Sep 30 '15 at 0:40
  • $\begingroup$ I view the molecules as scattered over all of space, or at least in a region wnormously larger than a ball of radius $R$. $\endgroup$ – André Nicolas Sep 30 '15 at 0:41
  • 1
    $\begingroup$ I agree that it is a reasonable model to postulate at the beginning of a problem. The OP's problem is already established. Why should we assume that their model is this one? $\endgroup$ – Ian Sep 30 '15 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.