Expectation of k-th order statistic of Negative Binomial Distribution
$begingroup$
Let $ X_1, X_2,...$ be i.i.d $NB(k,q)$. I am interested in calculating the expectation of their k-th order statistic $X_{k:n}$. From my understanding of order statistics, the CDF of $X_{k:n}$ is given as
$$ P[X_{k:n} leq x] = sum_{j=k}^n {n choose j} (F(x))^j(1-F(x))^{n-j}$$
From CDF, the PMF can be derived as:
$$P[X_{k:n} = x] = P[X_{k:n} leq x] - P[X_{k:n} leq x-1] $$
and the expectation :
$$E[X_{k:n}] = sum_{x = k}^infty x P[X_{k:n} = x]$$
I fear the direct calculation in such manner would involve complex algebra. Is there any simple way to derive such expectation. I also read the "The Order Statistics of the Negative Binomial Distribution" by D. H. Young (link to the paper - The Order Statistics of the Negative Binomial Distribution). Although, the author uses the recurrence relation to calculate m-th moments.
Any help in this matter, even a simple algebra trick would be much appreciated.
order-statistics negative-binomial
asked Dec 3 '18 at 0:21
Resting PlatypusResting Platypus
12
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add a comment |
$begingroup$
Let $ X_1, X_2,...$ be i.i.d $NB(k,q)$. I am interested in calculating the expectation of their k-th order statistic $X_{k:n}$. From my understanding of order statistics, the CDF of $X_{k:n}$ is given as
$$ P[X_{k:n} leq x] = sum_{j=k}^n {n choose j} (F(x))^j(1-F(x))^{n-j}$$
From CDF, the PMF can be derived as:
$$P[X_{k:n} = x] = P[X_{k:n} leq x] - P[X_{k:n} leq x-1] $$
and the expectation :
$$E[X_{k:n}] = sum_{x = k}^infty x P[X_{k:n} = x]$$
I fear the direct calculation in such manner would involve complex algebra. Is there any simple way to derive such expectation. I also read the "The Order Statistics of the Negative Binomial Distribution" by D. H. Young (link to the paper - The Order Statistics of the Negative Binomial Distribution). Although, the author uses the recurrence relation to calculate m-th moments.
Any help in this matter, even a simple algebra trick would be much appreciated.
order-statistics negative-binomial
asked Dec 3 '18 at 0:21
Resting PlatypusResting Platypus
12
$endgroup$
add a comment |
$begingroup$
Let $ X_1, X_2,...$ be i.i.d $NB(k,q)$. I am interested in calculating the expectation of their k-th order statistic $X_{k:n}$. From my understanding of order statistics, the CDF of $X_{k:n}$ is given as
$$ P[X_{k:n} leq x] = sum_{j=k}^n {n choose j} (F(x))^j(1-F(x))^{n-j}$$
From CDF, the PMF can be derived as:
$$P[X_{k:n} = x] = P[X_{k:n} leq x] - P[X_{k:n} leq x-1] $$
and the expectation :
$$E[X_{k:n}] = sum_{x = k}^infty x P[X_{k:n} = x]$$
I fear the direct calculation in such manner would involve complex algebra. Is there any simple way to derive such expectation. I also read the "The Order Statistics of the Negative Binomial Distribution" by D. H. Young (link to the paper - The Order Statistics of the Negative Binomial Distribution). Although, the author uses the recurrence relation to calculate m-th moments.
Any help in this matter, even a simple algebra trick would be much appreciated.
order-statistics negative-binomial
asked Dec 3 '18 at 0:21
Resting PlatypusResting Platypus
12
$endgroup$
Let $ X_1, X_2,...$ be i.i.d $NB(k,q)$. I am interested in calculating the expectation of their k-th order statistic $X_{k:n}$. From my understanding of order statistics, the CDF of $X_{k:n}$ is given as
$$ P[X_{k:n} leq x] = sum_{j=k}^n {n choose j} (F(x))^j(1-F(x))^{n-j}$$
From CDF, the PMF can be derived as:
$$P[X_{k:n} = x] = P[X_{k:n} leq x] - P[X_{k:n} leq x-1] $$
and the expectation :
$$E[X_{k:n}] = sum_{x = k}^infty x P[X_{k:n} = x]$$
I fear the direct calculation in such manner would involve complex algebra. Is there any simple way to derive such expectation. I also read the "The Order Statistics of the Negative Binomial Distribution" by D. H. Young (link to the paper - The Order Statistics of the Negative Binomial Distribution). Although, the author uses the recurrence relation to calculate m-th moments.
Any help in this matter, even a simple algebra trick would be much appreciated.
order-statistics negative-binomial
order-statistics negative-binomial
asked Dec 3 '18 at 0:21
Resting PlatypusResting Platypus
12
asked Dec 3 '18 at 0:21
Resting PlatypusResting Platypus
12
asked Dec 3 '18 at 0:21
Resting PlatypusResting Platypus
12
asked Dec 3 '18 at 0:21
Resting PlatypusResting Platypus
12
asked Dec 3 '18 at 0:21
Resting PlatypusResting Platypus
12
12
add a comment |
add a comment |
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