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Chapter 8: Q11 E (page 473)

If a random variableXhas the\({\chi ^2}\)distribution withmdegrees of freedom, then the distribution of\({X^{\frac{1}{2}}}\)is called achi(\(\chi \)) distribution with m degrees of freedom. Determinethe mean of this distribution.

Short Answer

Expert verified

\(E\left( {{X^{\frac{1}{2}}}} \right) = \frac{{\sqrt 2 \Gamma \left( {^{\frac{{n + 1}}{2}}} \right)}}{{\Gamma \left( {^{\frac{n}{2}}} \right)}}\)

Step by step solution

01

Given information

X follows \({\chi ^2}\)a distribution with m degrees of freedom.

02

Calculate the mean 

\(\begin{align}E\left( {{X^{\frac{1}{2}}}} \right) &= \int\limits_0^\infty {{x^{^{\frac{1}{2}}}}\frac{1}{{{2^{\frac{n}{2}}}\Gamma \left( {^{\frac{n}{2}}} \right)}}} {x^{^{\frac{n}{2} - 1}}}{e^{{ - ^{\frac{x}{2}}}}}dx\\ &= \frac{1}{{{2^{\frac{n}{2}}}\Gamma \left( {^{\frac{n}{2}}} \right)}}\int\limits_0^\infty {{x^{^{\frac{{n - 1}}{2}}}}{e^{{ - ^{\frac{x}{2}}}}}dx} \\ &= \frac{1}{{{2^{\frac{n}{2}}}\Gamma \left( {^{\frac{n}{2}}} \right)}}{2^{^{\frac{{n + 1}}{2}}}}\Gamma \left( {^{\frac{{n + 1}}{2}}} \right)\\ &= \frac{{\sqrt 2 \Gamma \left( {^{\frac{{n + 1}}{2}}} \right)}}{{\Gamma \left( {^{\frac{n}{2}}} \right)}}\end{align}\)

Hence, \(E\left( {{X^{\frac{1}{2}}}} \right) = \frac{{\sqrt 2 \Gamma \left( {^{\frac{{n + 1}}{2}}} \right)}}{{\Gamma \left( {^{\frac{n}{2}}} \right)}}\).

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Most popular questions from this chapter

For the conditions of Exercise 5, use the central limit theorem in Sec. 6.3 to find approximately the size of a random sample that must be taken so that \(P\left( {{\bf{|}}{{{\bf{\bar X}}}_{\bf{n}}} - {\bf{p|}}} \right) \ge 0.95\) whenp=0.2.

Question:Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from a distribution for which the p.d.f. or the p.f. is f (x|胃 ), where the value of the parameter 胃 is unknown. Let\({\bf{X = }}\left( {{{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}} \right)\)and let T be a statistic. Assuming that 未(X) is an unbiased estimator of 胃, it does not depend on 胃. (If T is a sufficient statistic defined in Sec. 7.7, then this will be true for every estimator 未. The condition also holds in other examples.) Let\({{\bf{\delta }}_{\bf{0}}}\left( {\bf{T}} \right)\)denote the conditional mean of 未(X) given T.

a. Show that\({{\bf{\delta }}_{\bf{0}}}\left( {\bf{T}} \right)\)is also an unbiased estimator of 胃.

b. Show that\({\bf{Va}}{{\bf{r}}_{\bf{\theta }}}\left( {{{\bf{\delta }}_{\bf{0}}}} \right) \le {\bf{Va}}{{\bf{r}}_{\bf{\theta }}}\left( {\bf{\delta }} \right)\)for every possible value of 胃. Hint: Use the result of Exercise 11 in Sec. 4.7.

Suppose that \({X_1},...,{X_n}\) form a random sample from the Bernoulli distribution with parameter p. Let \({\bar X_n}\) be the sample average. Use the variance stabilizing transformation found in Exercise 5 of Section 6.5 to construct an approximate coefficient 纬 confidence interval for p

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{, \ldots ,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the normal distribution with mean渭and variance\({{\bf{\sigma }}^{\bf{2}}}\), and let\({{\bf{\hat \sigma }}^{\bf{2}}}\)denote the sample variance. Determine the smallest values ofnfor which the following relations are satisfied:

  1. \({\bf{Pr}}\left( {\frac{{{{{\bf{\hat \sigma }}}^{\bf{2}}}}}{{{{\bf{\sigma }}^{\bf{2}}}}} \le {\bf{1}}{\bf{.5}}} \right) \ge {\bf{0}}{\bf{.95}}\)
  2. \({\bf{Pr}}\left( {\left| {{{{\bf{\hat \sigma }}}^{\bf{2}}}{\bf{ - }}{{\bf{\sigma }}^{\bf{2}}}} \right| \le \frac{{\bf{1}}}{{\bf{2}}}{{\bf{\sigma }}^{\bf{2}}}} \right) \ge {\bf{0}}{\bf{.8}}\)

Question: Consider the situation described in Exercise 7 of Sec. 8.5. Use a prior distribution from the normal-gamma family with values \({{\bf{\mu }}_{\bf{0}}}{\bf{ = 150,}}{{\bf{\lambda }}_{\bf{0}}}{\bf{ = 0}}{\bf{.5,}}{{\bf{\alpha }}_{\bf{0}}}{\bf{ = 1,}}{{\bf{\beta }}_{\bf{0}}}{\bf{ = 4}}\)

a. Find the posterior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau = }}\frac{{\bf{1}}}{{{{\bf{\sigma }}^{\bf{2}}}}}\)

b. Find an interval (a, b) such that the posterior probability is 0.90 that a <\({\bf{\mu }}\)<b.
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