Moments Of Negative Binomial Distribution
Moments of negative binomial distribution Let X follows a binomial distribution i.e X ~ B- ( k, p ) or X ~ NB ( k, p ) where X is a random variable ( r.v ) and k and p are the parameters. Then, the rthmoment about origin of X is given by $$μ_r^| = E (x) \ = \sum_{x=0}^\infty x^r p(x)$$ $$= \sum_{x=0}^\infty x^r \binom{x+k-1}{k-1} p^k q^x$$ Now, putting r = 1, we get, $$μ_1^| = E (x) \ = \sum_{x=0}^\infty x \binom{x+k-1}{k-1} p^k q^x$$ $$= \sum_{x=0}^\infty \frac{(x+k-1)!}{(x-1)! (k-1)!} p^K q^x$$ $$= \sum_{x=1}^\infty \frac{kq}{p} \binom{x+k-1}{k}p^k+1 q^x-1 $$ $$= \frac{kq}{p} $$∴ \ Mean \ of \ negative \ binomial \ distribution \ is \frac{kq}{p}$$ Putting r = 2, we get, μ2′= E (x2) $$= \sum_{x=0}^\infty x^2\binom{x+k-1}{k-1}p^kq^x$$ $$= \sum_{x=0}^\infty {x (x - 1) + x} \ \binom{x+k-1}{k-1} p^kq^x$$ $$= \sum_{x=0}^\infty \frac{k(k+1)q^2}{p^2} \binom{x+k-1}{k-1} \ {p^k}^{+2} {q^x}{-2} \ + \sum_{x=1}^\infty \frac{kq}{p} \binom{x+k-1}{k-1} \ {p^k}^{+1} {q^x}^{-1}$$ $$= \frac{k (k + 1)q^2}{p^2} \ + \ \frac{kq}{p}$$ ∴ V (x) =μ2 =μ2′ - (μ1′)2 $$= \frac{k(k+1)q^2}{p^2} + \frac{kq}{p} - \frac{k^2q^2}{p^2}$$ $$= \frac{kq}{p^2}$$ ∴ Variance of negative binomial distribution is $$V(x)= \frac{kq}{p^2}$$ $$As \ k \ > \ 1, \ p \ < \ 1 \ we \ have \ \frac{kq}{p} \ < \frac{kq}{p^2}$$ i.e. mean < variance which is an unique property of negative binomial distribution. The higher order moments of negative binomial distribution can be obtained on putting r = 3 and 4. Moment generating function of negative binomial distribution Let x followsnegative binomial distribution with parameters k and p. Then the probability mass function of X is $$ p(x) \ = \binom{x+k-1}{k-1} p^kq^x$$ $$If \ p \ = \frac{1}{Q} \ and \ q \ = \frac{P}{Q}, \ we \ get$$ $$ p(x) = \binom{x+k-1}{k-1} \ {Q^-}^{k} \ \frac{P^x}{Q^x}$$ The moment generating function of X ~ B- (k, p) is Mx(t) = E (etx) $$= \sum_{x=0}^\infty {e^t}{x} p(x)$$ $$= \sum_{x=0}^\infty {e^t}^{x} \binom{x+k-1}{k-1} {Q^-}^{k} \frac{P^x}{Q^x}$$ $$= \sum_{x=0}^\infty \binom{x+k-1}{k-1} {Q^-}^{k} \frac{(Pe^t)^x}{Q^x}$$ $$= \ {Q^-}^{k} \sum_{x=0}^\infty \binom{x+k-1}{k-1} \frac{(Pe^t)^x}{Q^x}$$ = (Q - Pet)-k ∴ Mx(t) =(Q - Pet)-k Now, the rth moment about origin of negative binomial distribution can be obtained by $$\mu_{r}^{'} \ = \frac{d^r M_x(t)}{dt^r} |_{t=0}$$ $$\therefore \mu_{1}^{'} \ = \frac{d M_x(t)}{dt^r} |_{t=0}$$ $$= \frac{d}{dt} (Q - Pe^t)^{-k} |_{t=0}$$ $$= \ (-k) \ (Q - Pe^t)^{-k-1} \ (-Pe^t) |_{t=0}$$ $$= kP$$ $$\therefore Mean, \ \mu_{1}^{'} \ = \ KP$$ Hence, the mean of negative binomial distribution is KP. $$Since, \ p \ = \frac{1}{Q} \Rightarrow \ Q \ = \frac{p}{Q} \ and \ q = \frac{P}{Q} \ \Rightarrow \ P \ = \ qQ \ = \frac{q}{p}$$ $$\mu_{1}^{'} \ = \ KP \ = \frac{kq}{p}$$ Similarly $$\mu_{2}{"} = \frac{d^2 M_x(t)}{dt^2} |_{t=0}$$ $$= \frac{d}{dt} \ {kPe^t (Q - Pe^t)^(-k-1)}|_{t=0}$$ $$ = \ kPe^t (Q- Pe^t)^(-k-1) \ + \ (-k-1) \ kPe^t (Q- Pe^t)^(-k-2) \ (-Pe^t)|_{t=0}$$ $$= \ kP \ + \ k(k+1) P^2$$ We have, $$\mu_{2} \ = \mu_{2}^{'} \ - (\mu_{1}^{'})^2$$ $$= \ kP \ + \ k (k + 1) P^2 \ - (kP)^2$$ $$= \ kP \ + \ kP^2$$ $$= \ kPQ$$ $$\therefore Variance, \mu_{2} \ = \ kPQ$$ Hence, variance of a negative binomial distribution is kPQ. $$Also, \ V(x) \ = \mu_{2} \ = \ = \ kPQ \ = \frac{kq}{p^2}$$ As Q > 1, kP < KPQ i.e. mean < variance. Thus, it is important to note that for negative binomial distribution. Mean < Variance The other higher moments can be obtained in the same manner from the moment generating function Mx(t). Converting P and Q in terms of p and q, we get $$ \mu_{3} \ = \frac{kq(1+p)}{p^3} \ and \mu_{4} \ = \frac{kq \ [p^2 + 3q (k+2)] \ }{p^4}$$ Then, moment coefficient of skeweness is $$\beta_{1} = \frac{(1+q)^2}{kq}$$ $$\gamma_{1} = \frac{ 1+ q} {\sqrt{kq}}$$ The moment coefficient of kurtosis is $$\beta_{2} \ = \frac{p^2 + 3q(k+1)}{kq}$$ $$\gamma_{2} \ = \frac{p^2 + 6q}{kq}$$ Additative property of negative binomial distribution Let X1, X2, ..., Xn be independent NB (ki, p) random variables i = 0, 1, 2, ..., n, respectively. Then we have Mxi(t) = (Q- Pet)-ki $$Now, \ the \ moment \ generating \ function \ of \ sum \ S_n \ = \sum_{x=0}^n X_i \ is \ given \ by$$ Msn(t) = MΣxi(t) = Mx1(t) . Mx2(t) ... Mxn(t) =(Q - Pet)-k1 (Q - Pet)-k2 ... (Q - Pet)-kn = (Q - Pet)-(k1+ k2 + ... + kn) which is the moment generating function of negative binomial distribution with parameters k1+ k2 + ... + kn and p.Hence by uniqueness theorem of mgfs, $$the \ sum \ S_n \ = \sum_{i=0}^n \ X_i \ has \ negative \ binomial \ distribution \ with \ parameters \sum_{i=0}^n k_i \ and \ p.$$ Recurrence relation for negative binomial distribution ( fitting of NBD) : Let X follows a negative binomial distribution i.e. X ~ B- (k, p). Then we have $$p (x) = \binom{x+k-1}{k-1} \ p^k \ q^x$$ $$and \ p(x+1) \ = \binom{x+k}{k-1} \ p^k \ {q^x}^{+1}$$ Now, $$\frac{p(x+1)}{p(x)} \ = \frac{(x+k)! (k-1)! x!}{(x+1)! (k-1)! (x+k-1)!} \ . \ q$$ $$= \frac{
Summary
Moments of negative binomial distribution Let X follows a binomial distribution i.e X ~ B- ( k, p ) or X ~ NB ( k, p ) where X is a random variable ( r.v ) and k and p are the parameters. Then, the rthmoment about origin of X is given by $$μ_r^| = E (x) \ = \sum_{x=0}^\infty x^r p(x)$$ $$= \sum_{x=0}^\infty x^r \binom{x+k-1}{k-1} p^k q^x$$ Now, putting r = 1, we get, $$μ_1^| = E (x) \ = \sum_{x=0}^\infty x \binom{x+k-1}{k-1} p^k q^x$$ $$= \sum_{x=0}^\infty \frac{(x+k-1)!}{(x-1)! (k-1)!} p^K q^x$$ $$= \sum_{x=1}^\infty \frac{kq}{p} \binom{x+k-1}{k}p^k+1 q^x-1 $$ $$= \frac{kq}{p} $$∴ \ Mean \ of \ negative \ binomial \ distribution \ is \frac{kq}{p}$$ Putting r = 2, we get, μ2′= E (x2) $$= \sum_{x=0}^\infty x^2\binom{x+k-1}{k-1}p^kq^x$$ $$= \sum_{x=0}^\infty {x (x - 1) + x} \ \binom{x+k-1}{k-1} p^kq^x$$ $$= \sum_{x=0}^\infty \frac{k(k+1)q^2}{p^2} \binom{x+k-1}{k-1} \ {p^k}^{+2} {q^x}{-2} \ + \sum_{x=1}^\infty \frac{kq}{p} \binom{x+k-1}{k-1} \ {p^k}^{+1} {q^x}^{-1}$$ $$= \frac{k (k + 1)q^2}{p^2} \ + \ \frac{kq}{p}$$ ∴ V (x) =μ2 =μ2′ - (μ1′)2 $$= \frac{k(k+1)q^2}{p^2} + \frac{kq}{p} - \frac{k^2q^2}{p^2}$$ $$= \frac{kq}{p^2}$$ ∴ Variance of negative binomial distribution is $$V(x)= \frac{kq}{p^2}$$ $$As \ k \ > \ 1, \ p \ < \ 1 \ we \ have \ \frac{kq}{p} \ < \frac{kq}{p^2}$$ i.e. mean < variance which is an unique property of negative binomial distribution. The higher order moments of negative binomial distribution can be obtained on putting r = 3 and 4. Moment generating function of negative binomial distribution Let x followsnegative binomial distribution with parameters k and p. Then the probability mass function of X is $$ p(x) \ = \binom{x+k-1}{k-1} p^kq^x$$ $$If \ p \ = \frac{1}{Q} \ and \ q \ = \frac{P}{Q}, \ we \ get$$ $$ p(x) = \binom{x+k-1}{k-1} \ {Q^-}^{k} \ \frac{P^x}{Q^x}$$ The moment generating function of X ~ B- (k, p) is Mx(t) = E (etx) $$= \sum_{x=0}^\infty {e^t}{x} p(x)$$ $$= \sum_{x=0}^\infty {e^t}^{x} \binom{x+k-1}{k-1} {Q^-}^{k} \frac{P^x}{Q^x}$$ $$= \sum_{x=0}^\infty \binom{x+k-1}{k-1} {Q^-}^{k} \frac{(Pe^t)^x}{Q^x}$$ $$= \ {Q^-}^{k} \sum_{x=0}^\infty \binom{x+k-1}{k-1} \frac{(Pe^t)^x}{Q^x}$$ = (Q - Pet)-k ∴ Mx(t) =(Q - Pet)-k Now, the rth moment about origin of negative binomial distribution can be obtained by $$\mu_{r}^{'} \ = \frac{d^r M_x(t)}{dt^r} |_{t=0}$$ $$\therefore \mu_{1}^{'} \ = \frac{d M_x(t)}{dt^r} |_{t=0}$$ $$= \frac{d}{dt} (Q - Pe^t)^{-k} |_{t=0}$$ $$= \ (-k) \ (Q - Pe^t)^{-k-1} \ (-Pe^t) |_{t=0}$$ $$= kP$$ $$\therefore Mean, \ \mu_{1}^{'} \ = \ KP$$ Hence, the mean of negative binomial distribution is KP. $$Since, \ p \ = \frac{1}{Q} \Rightarrow \ Q \ = \frac{p}{Q} \ and \ q = \frac{P}{Q} \ \Rightarrow \ P \ = \ qQ \ = \frac{q}{p}$$ $$\mu_{1}^{'} \ = \ KP \ = \frac{kq}{p}$$ Similarly $$\mu_{2}{"} = \frac{d^2 M_x(t)}{dt^2} |_{t=0}$$ $$= \frac{d}{dt} \ {kPe^t (Q - Pe^t)^(-k-1)}|_{t=0}$$ $$ = \ kPe^t (Q- Pe^t)^(-k-1) \ + \ (-k-1) \ kPe^t (Q- Pe^t)^(-k-2) \ (-Pe^t)|_{t=0}$$ $$= \ kP \ + \ k(k+1) P^2$$ We have, $$\mu_{2} \ = \mu_{2}^{'} \ - (\mu_{1}^{'})^2$$ $$= \ kP \ + \ k (k + 1) P^2 \ - (kP)^2$$ $$= \ kP \ + \ kP^2$$ $$= \ kPQ$$ $$\therefore Variance, \mu_{2} \ = \ kPQ$$ Hence, variance of a negative binomial distribution is kPQ. $$Also, \ V(x) \ = \mu_{2} \ = \ = \ kPQ \ = \frac{kq}{p^2}$$ As Q > 1, kP < KPQ i.e. mean < variance. Thus, it is important to note that for negative binomial distribution. Mean < Variance The other higher moments can be obtained in the same manner from the moment generating function Mx(t). Converting P and Q in terms of p and q, we get $$ \mu_{3} \ = \frac{kq(1+p)}{p^3} \ and \mu_{4} \ = \frac{kq \ [p^2 + 3q (k+2)] \ }{p^4}$$ Then, moment coefficient of skeweness is $$\beta_{1} = \frac{(1+q)^2}{kq}$$ $$\gamma_{1} = \frac{ 1+ q} {\sqrt{kq}}$$ The moment coefficient of kurtosis is $$\beta_{2} \ = \frac{p^2 + 3q(k+1)}{kq}$$ $$\gamma_{2} \ = \frac{p^2 + 6q}{kq}$$ Additative property of negative binomial distribution Let X1, X2, ..., Xn be independent NB (ki, p) random variables i = 0, 1, 2, ..., n, respectively. Then we have Mxi(t) = (Q- Pet)-ki $$Now, \ the \ moment \ generating \ function \ of \ sum \ S_n \ = \sum_{x=0}^n X_i \ is \ given \ by$$ Msn(t) = MΣxi(t) = Mx1(t) . Mx2(t) ... Mxn(t) =(Q - Pet)-k1 (Q - Pet)-k2 ... (Q - Pet)-kn = (Q - Pet)-(k1+ k2 + ... + kn) which is the moment generating function of negative binomial distribution with parameters k1+ k2 + ... + kn and p.Hence by uniqueness theorem of mgfs, $$the \ sum \ S_n \ = \sum_{i=0}^n \ X_i \ has \ negative \ binomial \ distribution \ with \ parameters \sum_{i=0}^n k_i \ and \ p.$$ Recurrence relation for negative binomial distribution ( fitting of NBD) : Let X follows a negative binomial distribution i.e. X ~ B- (k, p). Then we have $$p (x) = \binom{x+k-1}{k-1} \ p^k \ q^x$$ $$and \ p(x+1) \ = \binom{x+k}{k-1} \ p^k \ {q^x}^{+1}$$ Now, $$\frac{p(x+1)}{p(x)} \ = \frac{(x+k)! (k-1)! x!}{(x+1)! (k-1)! (x+k-1)!} \ . \ q$$ $$= \frac{
Things to Remember
- In negative binomial distribution, number of trials is random variable and number of successes is fixed.
- $$Mean \ of \ negative \ binomial \ distribution \ is \ \frac{kq}{p}$$
- variance of negative binomial distribuyion is $$\frac{kq}{P^2}.$$
- In negative binomial distribution mean < variance.
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Moments Of Negative Binomial Distribution
Moments of negative binomial distribution
Let X follows a binomial distribution i.e X ~ B- ( k, p ) or X ~ NB ( k, p ) where X is a random variable ( r.v ) and k and p are the parameters. Then, the rthmoment about origin of X is given by
$$μ_r^| = E (x) \ = \sum_{x=0}^\infty x^r p(x)$$
$$= \sum_{x=0}^\infty x^r \binom{x+k-1}{k-1} p^k q^x$$
Now, putting r = 1, we get,
$$μ_1^| = E (x) \ = \sum_{x=0}^\infty x \binom{x+k-1}{k-1} p^k q^x$$
$$= \sum_{x=0}^\infty \frac{(x+k-1)!}{(x-1)! (k-1)!} p^K q^x$$
$$= \sum_{x=1}^\infty \frac{kq}{p} \binom{x+k-1}{k}p^k+1 q^x-1 $$
$$= \frac{kq}{p}
$$∴ \ Mean \ of \ negative \ binomial \ distribution \ is \frac{kq}{p}$$
Putting r = 2, we get,
μ2′= E (x2)
$$= \sum_{x=0}^\infty x^2\binom{x+k-1}{k-1}p^kq^x$$
$$= \sum_{x=0}^\infty {x (x - 1) + x} \ \binom{x+k-1}{k-1} p^kq^x$$
$$= \sum_{x=0}^\infty \frac{k(k+1)q^2}{p^2} \binom{x+k-1}{k-1} \ {p^k}^{+2} {q^x}{-2} \ + \sum_{x=1}^\infty \frac{kq}{p} \binom{x+k-1}{k-1} \ {p^k}^{+1} {q^x}^{-1}$$
$$= \frac{k (k + 1)q^2}{p^2} \ + \ \frac{kq}{p}$$
∴ V (x) =μ2 =μ2′ - (μ1′)2
$$= \frac{k(k+1)q^2}{p^2} + \frac{kq}{p} - \frac{k^2q^2}{p^2}$$
$$= \frac{kq}{p^2}$$
∴ Variance of negative binomial distribution is
$$V(x)= \frac{kq}{p^2}$$
$$As \ k \ > \ 1, \ p \ < \ 1 \ we \ have \ \frac{kq}{p} \ < \frac{kq}{p^2}$$
i.e. mean < variance which is an unique property of negative binomial distribution.
The higher order moments of negative binomial distribution can be obtained on putting r = 3 and 4.
Moment generating function of negative binomial distribution
Let x followsnegative binomial distribution with parameters k and p. Then the probability mass function of X is
$$ p(x) \ = \binom{x+k-1}{k-1} p^kq^x$$
$$If \ p \ = \frac{1}{Q} \ and \ q \ = \frac{P}{Q}, \ we \ get$$
$$ p(x) = \binom{x+k-1}{k-1} \ {Q^-}^{k} \ \frac{P^x}{Q^x}$$
The moment generating function of X ~ B- (k, p) is
Mx(t) = E (etx)
$$= \sum_{x=0}^\infty {e^t}{x} p(x)$$
$$= \sum_{x=0}^\infty {e^t}^{x} \binom{x+k-1}{k-1} {Q^-}^{k} \frac{P^x}{Q^x}$$
$$= \sum_{x=0}^\infty \binom{x+k-1}{k-1} {Q^-}^{k} \frac{(Pe^t)^x}{Q^x}$$
$$= \ {Q^-}^{k} \sum_{x=0}^\infty \binom{x+k-1}{k-1} \frac{(Pe^t)^x}{Q^x}$$
= (Q - Pet)-k
∴ Mx(t) =(Q - Pet)-k
Now, the rth moment about origin of negative binomial distribution can be obtained by
$$\mu_{r}^{'} \ = \frac{d^r M_x(t)}{dt^r} |_{t=0}$$
$$\therefore \mu_{1}^{'} \ = \frac{d M_x(t)}{dt^r} |_{t=0}$$
$$= \frac{d}{dt} (Q - Pe^t)^{-k} |_{t=0}$$
$$= \ (-k) \ (Q - Pe^t)^{-k-1} \ (-Pe^t) |_{t=0}$$
$$= kP$$
$$\therefore Mean, \ \mu_{1}^{'} \ = \ KP$$
Hence, the mean of negative binomial distribution is KP.
$$Since, \ p \ = \frac{1}{Q} \Rightarrow \ Q \ = \frac{p}{Q} \ and \ q = \frac{P}{Q} \ \Rightarrow \ P \ = \ qQ \ = \frac{q}{p}$$
$$\mu_{1}^{'} \ = \ KP \ = \frac{kq}{p}$$
Similarly
$$\mu_{2}{"} = \frac{d^2 M_x(t)}{dt^2} |_{t=0}$$
$$= \frac{d}{dt} \ {kPe^t (Q - Pe^t)^(-k-1)}|_{t=0}$$
$$ = \ kPe^t (Q- Pe^t)^(-k-1) \ + \ (-k-1) \ kPe^t (Q- Pe^t)^(-k-2) \ (-Pe^t)|_{t=0}$$
$$= \ kP \ + \ k(k+1) P^2$$
We have,
$$\mu_{2} \ = \mu_{2}^{'} \ - (\mu_{1}^{'})^2$$
$$= \ kP \ + \ k (k + 1) P^2 \ - (kP)^2$$
$$= \ kP \ + \ kP^2$$
$$= \ kPQ$$
$$\therefore Variance, \mu_{2} \ = \ kPQ$$
Hence, variance of a negative binomial distribution is kPQ.
$$Also, \ V(x) \ = \mu_{2} \ = \ = \ kPQ \ = \frac{kq}{p^2}$$
As Q > 1, kP < KPQ i.e. mean < variance. Thus, it is important to note that for negative binomial distribution.
Mean < Variance
The other higher moments can be obtained in the same manner from the moment generating function Mx(t).
Converting P and Q in terms of p and q, we get
$$ \mu_{3} \ = \frac{kq(1+p)}{p^3} \ and \mu_{4} \ = \frac{kq \ [p^2 + 3q (k+2)] \ }{p^4}$$
Then, moment coefficient of skeweness is
$$\beta_{1} = \frac{(1+q)^2}{kq}$$
$$\gamma_{1} = \frac{ 1+ q} {\sqrt{kq}}$$
The moment coefficient of kurtosis is
$$\beta_{2} \ = \frac{p^2 + 3q(k+1)}{kq}$$
$$\gamma_{2} \ = \frac{p^2 + 6q}{kq}$$
Additative property of negative binomial distribution
Let X1, X2, ..., Xn be independent NB (ki, p) random variables i = 0, 1, 2, ..., n, respectively. Then we have
Mxi(t) = (Q- Pet)-ki
$$Now, \ the \ moment \ generating \ function \ of \ sum \ S_n \ = \sum_{x=0}^n X_i \ is \ given \ by$$
Msn(t) = MΣxi(t)
= Mx1(t) . Mx2(t) ... Mxn(t)
=(Q - Pet)-k1 (Q - Pet)-k2 ... (Q - Pet)-kn
= (Q - Pet)-(k1+ k2 + ... + kn)
which is the moment generating function of negative binomial distribution with parameters k1+ k2 + ... + kn and p.Hence by uniqueness theorem of mgfs,
$$the \ sum \ S_n \ = \sum_{i=0}^n \ X_i \ has \ negative \ binomial \ distribution \ with \ parameters \sum_{i=0}^n k_i \ and \ p.$$
Recurrence relation for negative binomial distribution ( fitting of NBD) :
Let X follows a negative binomial distribution i.e. X ~ B- (k, p). Then we have
$$p (x) = \binom{x+k-1}{k-1} \ p^k \ q^x$$
$$and \ p(x+1) \ = \binom{x+k}{k-1} \ p^k \ {q^x}^{+1}$$
Now,
$$\frac{p(x+1)}{p(x)} \ = \frac{(x+k)! (k-1)! x!}{(x+1)! (k-1)! (x+k-1)!} \ . \ q$$
$$= \frac{x+k}{x+1} \ q$$
$$p(x+1) \ = (\frac{x+k}{x+1}) \ q \ p(x)$$
This is the recurrence relation for probabilities of negative binomial distribution for x= 0, 1, 2, ... and k > 0. This recurrence formula is useful for fitting the negative binomial distribution to the given empirical data. The excepted frequency of X is given by
fe (x) = N. p(x) for x = 0, 1, 2, ..., and k > 0
Where, N =Σfo, the sum of observed frequencies.
Bibliography
Sukubhattu N.P. (2013). Probability & Inference - II. Asmita Books Publishers & Distributors (P) Ltd., Kathmandu.
Larson H.J. Introduction to Probability Theory and Statistical Inference. WileyInternational, New York.
Lesson
Discrete probability distribution
Subject
Statistics
Grade
Bachelor of Science
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