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Chapter 5: Q10E (page 287)

Show that for all positive integersnandk,\[\left( {\begin{array}{*{20}{c}}{ - n}\\k\end{array}} \right) = {\left( { - 1} \right)^k}\left( {\begin{array}{*{20}{c}}{n + k - 1}\\k\end{array}} \right)\].

Short Answer

Expert verified

\[\left( {\begin{array}{*{20}{c}}{ - n}\\k\end{array}} \right) = {\left( { - 1} \right)^k}\left( {\begin{array}{*{20}{c}}{n + k - 1}\\k\end{array}} \right)\]

Step by step solution

01

Given information

n and k both are positive integers.

02

Compute the required value

Referring to the equations 5.3.13 and 5.3.14.

\[\begin{array}{c}\left( {\begin{array}{*{20}{c}}{\bf{m}}\\{\bf{r}}\end{array}} \right){\bf{ = }}\frac{{{\bf{m!}}}}{{{\bf{r!}}\left( {{\bf{m}} - {\bf{r}}} \right){\bf{!}}}}\\{\bf{ = }}\frac{{{\bf{m}}\left( {{\bf{m}} - {\bf{1}}} \right)...\left( {{\bf{m}} - {\bf{r + 1}}} \right)}}{{{\bf{r!}}}}\end{array}\]

From the above equation, the value of\[\left( {\begin{array}{*{20}{c}}{ - n}\\k\end{array}} \right)\]is given by,

After reversing the order of the factors in the numerator, it becomes,

\[\begin{array}{c}\left( {\begin{array}{*{20}{c}}{ - n}\\k\end{array}} \right) = \frac{{{{\left( { - 1} \right)}^k}\left( {n + k - 1} \right)\left( {n + k - 2} \right)...n}}{{k!}}\\ = {\left( { - 1} \right)^k}\left( {\begin{array}{*{20}{c}}{n + k - 1}\\k\end{array}} \right)\end{array}\]

Hence, proved.

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Most popular questions from this chapter

Suppose that the random variables \({{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}\) are independent and that \({{\bf{X}}_{\bf{i}}}\) has the Poisson distribution with mean \({{\bf{\lambda }}_{\bf{i}}}\left( {{\bf{i = 1, \ldots ,k}}} \right)\). Show that for each fixed positive integer n, the conditional distribution of the random Vector \({\bf{X = }}\left( {{{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}} \right)\), given that \(\sum\limits_{{\bf{i = 1}}}^{\bf{k}} {{{\bf{X}}_{\bf{i}}}{\bf{ = n}}} \) it is the multinomial distribution with parameters n and

\(\begin{array}{l}{\bf{p = }}\left( {{{\bf{p}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{p}}_{\bf{k}}}} \right){\bf{,}}\,{\bf{where}}\\{{\bf{p}}_{\bf{i}}}{\bf{ = }}\frac{{{{\bf{\lambda }}_{\bf{i}}}}}{{\sum\limits_{{\bf{j = 1}}}^{\bf{k}} {{{\bf{\lambda }}_{\bf{j}}}} }}\,{\bf{for}}\,{\bf{i = 1, \ldots ,k}}{\bf{.}}\end{array}\)

Suppose that n items are being tested simultaneously, the items are independent, and the length of life of each item has the exponential distribution with parameter\(\beta \).Determine the expected length of time until three items have failed. Hint: The required value is\(E\left( {{Y_1} + {Y_2} + {Y_3}} \right)\)in the notation of Theorem 5.7.11.

Suppose that a random variable\(X\)has the normal distribution with mean\(\mu \)and variance\({\sigma ^2}\). Determine the value of\(E\left[ {{{\left( {X - \mu } \right)}^{2n}}} \right]\)for\(n = 1,2....\)

Let \({{\bf{X}}_{{\bf{1,}}}}{\bf{ }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\)be i.i.d. random variables having thenormal distribution with mean \({\bf{\mu }}\) and variance\({{\bf{\sigma }}^{\bf{2}}}\). Define\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\frac{{\bf{1}}}{{\bf{n}}}\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{X}}_{\bf{i}}}} \) , the sample mean. In this problem, weshall find the conditional distribution of each \({{\bf{X}}_{\bf{i}}}\)given\(\overline {{{\bf{X}}_{\bf{n}}}} \).

a.Show that \({{\bf{X}}_{\bf{i}}}\)and\(\overline {{{\bf{X}}_{\bf{n}}}} \) have the bivariate normal distributionwith both means \({\bf{\mu }}\), variances\({{\bf{\sigma }}^{\bf{2}}}{\rm{ }}{\bf{and}}\,\,\frac{{{{\bf{\sigma }}^{\bf{2}}}}}{{\bf{n}}}\),and correlation\(\frac{{\bf{1}}}{{\sqrt {\bf{n}} }}\).

Hint: Let\({\bf{Y = }}\sum\limits_{{\bf{j}} \ne {\bf{i}}} {{{\bf{X}}_{\bf{j}}}} \).

Nowshow that Y and \({{\bf{X}}_{\bf{i}}}\) are independent normals and \({{\bf{X}}_{\bf{n}}}\)and \({{\bf{X}}_{\bf{i}}}\) are linear combinations of Y and \({{\bf{X}}_{\bf{i}}}\) .

b.Show that the conditional distribution of \({{\bf{X}}_{\bf{i}}}\) given\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\overline {{{\bf{x}}_{\bf{n}}}} \)\(\) is normal with mean \(\overline {{{\bf{x}}_{\bf{n}}}} \) and variance \({{\bf{\sigma }}^{\bf{2}}}\left( {{\bf{1 - }}\frac{{\bf{1}}}{{\bf{n}}}} \right)\).

If a random sample of 25 observations is taken from the normal distribution with mean \(\mu \) and standard deviation 2, what is the probability that the sample mean will lie within one unit of 渭 ?
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