Question

Consider the probability function for the binomial distribution, $$ B(M)=\frac{N !}{M !(N-M) !} p^{M} q^{N-M}, $$ where \(N\) and \(p\) are specified, and the probability \(q=1-p\). a. Show that the distribution is properly normalized. b. Verify that the mean \(\bar{M}=N p\). c. Show that the standard deviation \(\sigma=\sqrt{N p(1-p)}\). d. The fluctuation is defined as the ratio of the standard deviation to the mean. Determine the fluctuation of the binomial distribution for \(p=0.5\). What happens to the fluctuation for large values of \(N\) ? What are the physical implications of this result when considering the thermodynamic properties of an ideal gas? Hint: Recall that, for arbitrary values of \(p\) and \(q\), the binomial theorem gives $$ \sum_{M=0}^{N} B(M)=(p+q)^{N} $$ so that, for any value of \(s\), $$ \sum_{M=0}^{N} s^{M} B(M)=(s p+q)^{N} . $$ The trick now is to differentiate with respect to \(s\).

Short answer

The binomial distribution is normalized since \(\sum_{M=0}^{N} B(M) = 1\). The mean is verified to be \(\bar{M}=N p\), and the standard deviation is \(\sigma=\sqrt{N p(1-p)}\). For \(p=0.5\), the fluctuation is \(\frac{1}{\sqrt{N}}\), which approaches 0 for large values of \(N\), indicating more predictable behavior for an ideal gas as the number of particles increases.

Step-by-step solution

Verify normalization of the distribution

To show that the distribution is properly normalized, we need to prove that the sum of probabilities for all possible outcomes is equal to 1. That is, we need to evaluate the sum \(\sum_{M=0}^{N} B(M)\) and show that it equals 1. Using the binomial theorem given in the hint, we set \(s=1\) which gives us the sum \(\sum_{M=0}^{N} B(M) = (1 * p + q)^N = (p+q)^{N}\). As \(q=1-p\), this simplifies to \((p + (1 - p))^N = 1^N = 1\).

Verify the mean of the distribution

To find the mean \(\bar{M}\), we need to evaluate \(\sum_{M=0}^{N} M B(M)\). Differentiating both sides of the equation \(\sum_{M=0}^{N} s^{M} B(M)=(s p+q)^{N}\) with respect to \(s\) gives us \(\sum_{M=0}^{N} M s^{M-1} B(M) = N (s p+q)^{N-1} p\). Setting \(s=1\), the left-hand side becomes the mean \(\bar{M}\) and the right-hand side simplifies to \(N p\). Thus, \(\bar{M} = N p\).

Verify the standard deviation of the distribution

To find the standard deviation \(\sigma\), we need to calculate the second moment \(\sum_{M=0}^{N} M^2 B(M)\) and use the definition \(\sigma = \sqrt{\bar{M^2} - (\bar{M})^2}\). Differentiating the equation obtained in Step 2 with respect to \(s\) again gives us \(\sum_{M=0}^{N} M (M-1) s^{M-2} B(M) = N(N-1) (s p+q)^{N-2} p^2\). Setting \(s=1\), we find that \(\bar{M^2} = N(N-1)p^2 + N p\). Plugging \(\bar{M^2}\) and \(\bar{M} = N p\) into the standard deviation formula provides \(\sigma = \sqrt{N p(1-p)}\).

Determine the fluctuation for p=0.5

The fluctuation is the ratio of the standard deviation to the mean. Using the results from the previous steps, this ratio for \(p=0.5\) is \(\frac{\sigma}{\bar{M}} = \frac{\sqrt{N p(1-p)}}{N p}\). Simplifying gives us \(\sqrt{\frac{1-p}{N p}}\) and substituting \(p=0.5\) yields \(\frac{1}{\sqrt{N}}\).

Analyze the fluctuation for large N

As \(N\) grows, the fluctuation \(\frac{1}{\sqrt{N}}\) approaches 0. This suggests that as the number of trials in the binomial distribution becomes very large, the relative standard deviation, compared to the mean, decreases. The physical implication for an ideal gas is that the deviation from the expected number of molecules having a particular property becomes less significant, indicating a more predictable behavior as the sample size increases.

Key concepts

Statistical Thermodynamics

Statistical thermodynamics is a branch of physics that combines laws of thermodynamics with statistical mechanics. It seeks to explain macroscopic properties of systems based on the behavior of their microscopic constituents. In essence, it enables scientists to calculate the macroscopic properties such as temperature, pressure, and entropy from the statistical properties of molecules and atoms.

In the context of the binomial distribution exercise, statistical thermodynamics would consider the distribution of a certain particle property (like energy levels) among a large number of particles in a system. For instance, if we consider an ideal gas where particles could be in two energy states, then the binomial distribution could model the probability of finding a certain number of particles in one of those states. Understanding the normalization and fluctuation within this distribution helps in predicting the thermodynamic properties of the system under study.

Distribution Normalization

Distribution normalization is a process of adjusting the statistical distribution so that the sum of all probabilities equals one. This process is critical because it ensures that the distribution can be interpreted as a proper probability, reflecting the certainty that one of the possible outcomes will occur.

In the given binomial distribution problem, normalization is verified by using the property that the sum of the probabilities over all possible outcomes (from 0 to N successes in N trials) is 1. This confirms that the distribution is properly normalized and can be used to calculate probabilities of different outcomes. In statistical thermodynamics, normalization is what allows us to associate probabilities with physical phenomena such as the likelihood of finding a system in a particular energy state.

Standard Deviation

The standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. In statistical terms, it's the square root of the average squared deviations from the mean.

In the context of binomial distribution, the standard deviation tells us how much the number of successes is expected to vary from the mean number of successes. For a binomial distribution with probability of success p, the standard deviation is given by \(\sigma = \sqrt{Np(1-p)}\), which provides a measure of the spread of the distribution around the mean \(\bar{M} = Np\). Understanding the standard deviation is essential in thermodynamics to predict the extent of fluctuations in properties such as pressure or volume for a thermodynamic system.

Fluctuation in Thermodynamics

In thermodynamics, fluctuation refers to the temporary and random deviation of a property from its average value. It's an intrinsic characteristic of systems that arises from the random motion of particles. The fluctuation is particularly relevant in statistical thermodynamics, where it explains the unpredictability of certain microscopic behaviors even if the macroscopic properties are stable.

The fluctuation ratio in binomial distribution is derived from the standard deviation \(\sigma\) and the mean \(\bar{M}\). For large values of N, the law of large numbers predicts that fluctuation decreases, which mirrors the behavior in large thermodynamic systems — individual fluctuations become insignificant, leading to predictable average properties. This relationship is crucial for understanding that in a sufficiently large system, certain macroscopic quantities like pressure and temperature become stable despite microscopic chaos.