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Show that the probability generating function of the negative multinomial distribution (I) with parameters \(\left(k ; p_{0}, p_{1}, \ldots, p_{r}\right)\) is $$ \varphi\left(t_{1}, \ldots, t_{r}\right)=p_{0}^{k}\left(1-\sum_{i=1}^{r} t_{i P i}\right)^{-k} $$

Short Answer

Expert verified
Successfully demonstrating the proof, the probability generating function of the negative multinomial distribution with parameters \((k ; p_{0}, p_{1}, \ldots, p_{r})\) is given by \(\varphi(t_{1}, ..., t_{r})=p_{0}^{k}\left(1-\sum_{i=1}^{r} t_{i} p_{i}\right)^{-k}\)

Step by step solution

01

Definition of the probability generating function (PGF)

Begin by writing down the generic form of a PGF. For a variable \(X\), the PGF, \(G_X(t)\) is defined as \(G_X(t) = E(t^X)\), the expected value of \(t\) raised to the power of \(X\).
02

Write the Negative Multinomial Distribution

Express the negative multinomial distribution as \(\left(k ; p_{0}, p_{1}, \ldots, p_{r}\right)\) where sum of the probabilities \(p_i = 1\) for \(i = 0, 1, ..., r\).
03

Express the PGF for Negative Multinomial Distribution

Apply the formula for the PGF to the negative multinomial distribution. The probability generating function of the negative multinomial distribution is the expectation of \((t_1^{X_1}... t_r^{X_r})\), which equals \(E((t_1^{X_1}... t_r^{X_r})) = \sum_{x_1 = 0}^{\infty}... \sum_{x_r = 0}^{\infty} (t_1^{x_1}... t_r^{x_r}) Pr((X_1 = x_1, ..., X_r = x_r))\)
04

Substitute the Negative Multinomial Distribution into the PGF

Substitute the negative multinomial distribution into the PGF resulting expression becomes \(\sum (t_1^{x_1}... t_r^{x_r}) Pr(X_1 = x_1, ..., X_r = x_r)\).
05

Simplification of the PGF Expression

Simplify. After summing and simplifying, one obtains the final expression as \(\varphi(t_{1}, ..., t_{r})=p_{0}^{k}\left(1-\sum_{i=1}^{r} t_{i} p_{i}\right)^{-k}\).

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Most popular questions from this chapter

The random variables \(X\) and \(Y\) have the following properties: \(X\) is positive, i.e., \(P\\{X>0\\}=1\), with continuous density funetion \(f(x)\), and \(Y \mid X\) has a uniform distribution on \(\\{0, X\\} .\) Prove: If \(Y\) and \(X-Y\) are independently dis* tributed, then $$ f(x)=a^{2} x e^{-a x}, \quad x>0, \quad a>0 $$

For each fixed \(\lambda>0\) let \(X\) have a Poisson distribution with parameter \(\lambda\). Suppose \(\lambda\) itself is a random variable following a gamma distribution (i.e., with density $$ f(\lambda)= \begin{cases}\frac{1}{\Gamma(n)} \lambda^{n-1} e^{-\lambda}, & \lambda \geq 0 \\ 0, & \lambda<0\end{cases} $$where \(n\) is a fixed positive constant). Show that now $$ \operatorname{Pr}\\{X=k\\}=\frac{\Gamma(k+n)}{\Gamma(n) \Gamma(k+1)}\left(\frac{1}{2}\right)^{k+n}, \quad k=0,1, \ldots $$ When \(n\) is an integer this is the negative binomial distribution with \(p=\frac{1}{2}\).

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