Question

Given that a particle is inside a sphere of radius \(1,\) and that it has equal probabilities of being found in any two volume elements of the same size, find the cumulative distribution function \(F(r)\) for the spherical coordinate \(r,\) and from it find the density function \(f(r) .\) Hint: \(F(r)\) is the probability that the particle is inside a sphere of radius \(r .\) Find \(\overline{r}\) and \(\sigma\).

Short answer

The CDF is \(F(r) = r^3\), the PDF is \(f(r) = 3r^2\), \(\overline{r} = \frac{3}{4}\), and \(\sigma = \frac{\sqrt{15}}{20}\).

Step-by-step solution

- Understanding the Problem

The particle is uniformly distributed inside a sphere of radius 1. This means any point within the sphere has an equal probability of having the particle.

- Calculating the Cumulative Distribution Function (CDF)

The Cumulative Distribution Function (CDF) is defined as the probability that the random variable, in this case, the radius, is less than or equal to a specific value. Calculate the volume of a sphere with radius r and divide it by the volume of the sphere with radius 1 (which is the total volume). Thus, \[ F(r) = \frac{\text{Volume of sphere with radius } r}{\text{Volume of sphere with radius } 1} = \frac{\frac{4}{3}\pi r^3}{\frac{4}{3} \pi (1)^3} = r^3 \] for \(0 \leq r \leq 1\).

- Determining the Density Function (PDF)

The density function, \(f(r)\), is the derivative of the cumulative distribution function. Find the derivative of \(F(r)\):\[ f(r) = \frac{dF(r)}{dr} = \frac{d}{dr}\big(r^3\big) = 3r^2 \].

- Expectation Value \(\overline{r}\)

The expectation value \(\overline{r}\) can be calculated using the integral of the product of the variable and its density function, from 0 to 1: \[ \overline{r} = \int_0^1 r f(r) dr = \int_0^1 r (3r^2) dr = 3 \int_0^1 r^3 dr = 3 \left[ \frac{r^4}{4} \right]_0^1 = 3 \cdot \frac{1}{4} = \frac{3}{4} \]

- Variance and Standard Deviation \(\sigma\)

First, calculate the expectation value of \(r^2\):\[ E(r^2) = \int_0^1 r^2 f(r) dr = \int_0^1 r^2 (3r^2) dr = 3 \int_0^1 r^4 dr = 3 \left[ \frac{r^5}{5} \right]_0^1 = 3 \cdot \frac{1}{5} = \frac{3}{5} \].Using the formula for variance, \(\sigma^2 = E(r^2) - (\overline{r})^2\):\[ \sigma^2 = \frac{3}{5} - \left( \frac{3}{4} \right)^2 = \frac{3}{5} - \frac{9}{16} = \frac{48}{80} - \frac{45}{80} = \frac{3}{80} \].Therefore, the standard deviation \(\sigma\) is:\[ \sigma = \sqrt{\frac{3}{80}} = \frac{\sqrt{3}}{4\sqrt{5}} = \frac{\sqrt{3}}{4\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{15}}{20} \].

Key concepts

Probability Density Function (PDF)

The probability density function (PDF) describes how the values of a random variable are distributed. The PDF is a function that shows the relative likelihood of a variable to take on a given value. In more technical terms, it is the derivative of the cumulative distribution function (CDF).
In our exercise, we need to find the PDF for the radius of a sphere where the particle is uniformly distributed. Since we already know the CDF, which is given by: \[ F(r) = r^3 \] for \[0 \leq r \leq 1\], we can find the PDF by differentiating the CDF: \[ f(r) = \frac{dF(r)}{dr} = \frac{d}{dr} \big( r^3 \big) = 3r^2 \].
This PDF, \( f(r) = 3r^2 \), tells us how the probability is distributed over the interval of \[r\] values. The graph of this PDF would show a quadratic relationship where the likelihood of the particle being at a certain radius increases with the square of the radius.

Uniform Distribution

A uniform distribution is a probability distribution where every outcome in a certain range is equally likely. In the context of our problem, the particle is uniformly distributed inside the sphere. This means the particle is just as likely to be found at any point within the sphere.
For a uniform distribution over a sphere with radius 1, every volume element of the sphere has the same probability density. Since the total volume of the sphere is \( \frac{4}{3} \pi \), the probability density function remains constant across the volume.
To translate this into our CDF and PDF, we determined that the CDF is \[ F(r) = \frac{4}{3} \pi r^3 / \frac{4}{3} \pi (1)^3 = r^3 \] and from this, the PDF is derived as \[ f(r) = 3r^2 \]. These formulas show the consistency of probability density across the sphere.

Variance and Standard Deviation

Variance and standard deviation are measures of the spread or dispersion of a set of values. In our exercise, they provide insight into how far the particle might be from the center on average.

First, we calculate the expectation value of \[ E(r) \] or mean radius, which we found was: \[ \overline{r} = \frac{3}{4} \].

Next, to find the variance \[ \sigma^2 \], we calculate the expectation value of \[ r^2 \], found to be \[ \frac{3}{5} \]. Using the variance formula: \[ \sigma^2 = E(r^2) - (\overline{r})^2 \], we substitute these values to find: \[ \sigma^2 = \frac{3}{5} - \left( \frac{3}{4} \right)^2 = \frac{3}{80} \].

Finally, the standard deviation \[ \sigma \] is the square root of \[ \sigma^2 \], calculated as: \[ \sigma = \sqrt{\frac{3}{80}} = \frac{\sqrt{15}}{20} \].
Understanding \[ \sigma \] helps predict how the particle is likely to spread inside the sphere, offering a quantifiable measure of deviation from the mean radius.