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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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Suppose that \({X_1},...,{X_n}\) form a random sample from the normal distribution with unknown mean μ (−∞ < μ < ∞) and known precision \(\tau \) . Suppose also that the prior distribution of μ is the normal distribution with mean \({\mu _0}\) and precision \({\lambda _0}\) . Show that the posterior distribution of μ, given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the normal distribution with mean

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