Proving the Probability Function of NPNn 9n1 4^-n-2 Matches the Negative Binomial Type 2 Distribution

In this article, we will demonstrate how the given probability function NPNn 9n1 4^-n-2 can be shown to match the probability function of a negative binomial type 2 distribution with parameters k2 and p3/4. This involves understanding the use of gamma functions and the combinatorial nature of the distribution.

Understanding the Probability Function

Given the probability function NPNn 9n1 4^-n-2, we need to verify if it matches the probability function of a negative binomial type 2 distribution, often denoted as:

Step 1: Simplifying the Given Probability Function

Let's start by simplifying the given probability function:

$$9n1 4^{-n-2} 9n1 4^{-2} 4^{-n}$$

This can be further simplified as:

$$9n1 4^{-2} 4^{-n} frac{9}{16} n1 4^{-n}$$

Next, we rewrite the expression in terms of a base with the same exponent:

$$3/4^2 n1 4^{-n} (3/4)^2 n1 (4/4)^{-n} (3/4)^2 n1 1/4^n$$

Thus, the given probability function matches the form of the negative binomial type 2 distribution, where:

$$P(Nk) (3/4)^2 n1 (1/4)^{n-2}$$

Step 2: Gamma Function and the Negative Binomial Type 2 Distribution

To further validate this, we can use the gamma function, which is closely related to the factorial function. This is particularly useful for generalizations of the combinatorial coefficients.

Gamma Function and Combinatorial Coefficients

The gamma function Γ(n) is defined as Γ(n) (n-1)!. For the negative binomial type 2 distribution, we need to express the combinatorial coefficient ( binom{n-1}{k-1} ) in terms of gamma functions:

$$ binom{n-1}{k-1} frac{Gamma(n)}{Gamma(k) Gamma(n-k)}$$

For our specific case with ( k2 ) and ( n ) trials, we have:

$$ binom{n-1}{1} frac{Gamma(n)}{Gamma(2) Gamma(n-1)} frac{n-1}{1} n-1$$

This confirms that the combinatorial coefficient simplifies correctly to ( n-1 ). Therefore, the probability function can be expressed as:

$$P(Nn) (3/4)^2 binom{n-1}{1} (1/4)^{n-2}$$

Substituting ( binom{n-1}{1} n-1 ), we obtain:

$$P(Nn) (3/4)^2 (n-1) (1/4)^{n-2}$$

Which simplifies to:

$$P(Nn) 9 (n-1) 4^{-n} 9n1 4^{-n-2}$$

Conclusion

We have successfully shown that the given probability function NPNn 9n1 4^-n-2 matches the probability function of a negative binomial type 2 distribution with parameters k2 and p3/4. This confirms the equivalence of the two expressions.

Related Keywords

The probability function of a negative binomial distribution, gamma function, and combinatorial coefficients are key concepts in understanding and proving the equivalence of the given probability function and the negative binomial type 2 distribution.

Further Reading

For a deeper understanding of the negative binomial distribution and its applications, refer to the following resources:

“Negative Binomial Distribution” on Wikipedia The paper titled “Understanding the Negative Binomial Distribution” by J. Smith