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Find the mode of theχ2 distribution withmdegrees of

freedom(m=1,2, . . .).

Short Answer

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Step by step solution

01

Given information

has a chi-square distribution with m degree of freedom\({\chi ^2}\)

02

The pdf of chi-square distribution

\(\)\(f\left( x \right) = {2^{ - m/2}}\Gamma \left( {m/2} \right){x^{m/2 - 1}}{e^{ - x/2}}\,;\,x > 0\)

03

Finding mode

To find the mode of the distribution we need to differentiate log f(x) with respect to x, and need to equate it with 0.

\(\)\(\begin{align}f\left( x \right) &= {2^{ - m/2}}\Gamma \left( {m/2} \right){x^{m/2 - 1}}{e^{ - x/2}}\,\\\log \,f\left( x \right) &= - \frac{m}{2}\log 2 - \log \Gamma \left( {\frac{m}{2}} \right) + \left( {\frac{m}{2} - 1} \right)\log x - \frac{x}{2}\\\frac{{d\log f\left( x \right)}}{{dx}} &= \left( {\frac{m}{2} - 1} \right)\frac{1}{x} - \frac{1}{2}\end{align}\)

\(\begin{align}Now,\,\,\frac{{d\log f\left( x \right)}}{{dx}} &= 0\\\left( {\frac{m}{2} - 1} \right)\frac{1}{x} - \frac{1}{2} &= 0\\\left( {\frac{m}{2} - 1} \right)\frac{1}{x} &= \frac{1}{2}\\\left( {\frac{{m - 2}}{2}} \right)\frac{1}{x} &= \frac{1}{2}\\x &= m - 2\end{align}\)

So the mode of the chi-square distribution is m-2 for \(m > 2\)

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Most popular questions from this chapter

Consider again the conditions of Exercise 19, and let\({{\bf{\hat \beta }}_{\bf{n}}}\)n denote the M.L.E. of β.

a. Use the delta method to determine the asymptotic distribution of\(\frac{{\bf{1}}}{{{{{\bf{\hat \beta }}}_{\bf{n}}}}}\).

b. Show that\(\frac{{\bf{1}}}{{{{{\bf{\hat \beta }}}_{\bf{n}}}}}{\bf{ = }}{{\bf{\bar X}}_{\bf{n}}}\), and use the central limit theorem to determine the asymptotic distribution of\(\frac{{\bf{1}}}{{{{{\bf{\hat \beta }}}_{\bf{n}}}}}\).

Question:Suppose that a random variable X has the geometric distribution with an unknown parameter p (0<p<1). Show that the only unbiased estimator of p is the estimator \({\bf{\delta }}\left( {\bf{X}} \right)\) such that \({\bf{\delta }}\left( {\bf{0}} \right){\bf{ = 1}}\) and \({\bf{\delta }}\left( {\bf{X}} \right){\bf{ = 0}}\) forX>0.

Suppose that \({X_1},...,{X_n}\) are i.i.d. having the normal distribution with mean μ and precision \(\tau \) given (μ, \(\tau \) ). Let (μ, \(\tau \) ) have the usual improper prior. Let \({\sigma '^{2}} = \frac{{s_n^2}}{{n - 1}}\) . Prove that the posterior distribution of \(V = \left( {n - 1} \right){\sigma '^{2}}\tau \) is the χ2 distribution with n − 1 degrees of freedom.

Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{, }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\) form a random sample from the normal distribution with unknown mean \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\), and also that the joint prior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\) is the normal-gamma distribution satisfying the following conditions: \({\bf{E}}\left( {\bf{\mu }} \right){\bf{ = 0}}\,\,\,\,{\bf{,E}}\left( {\bf{\tau }} \right){\bf{ = 2,E}}\left( {{{\bf{\tau }}^{\bf{2}}}} \right){\bf{ = 5}}\,\,\,{\bf{and}}\,\,{\bf{Pr}}\left( {\left| {\bf{\mu }} \right|{\bf{ < 1}}{\bf{.412}}} \right){\bf{ = 0}}{\bf{.5}}\)Determine the prior hyperparameters \({{\bf{\mu }}_{\bf{0}}}{\bf{,}}{{\bf{\lambda }}_{\bf{0}}}{\bf{,}}{{\bf{\alpha }}_{\bf{0}}}{\bf{,}}{{\bf{\beta }}_{\bf{0}}}\)

Show that two random variables \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\)cannot have the joint normal-gamma distribution such that

\({\bf{E}}\left( {\bf{\mu }} \right){\bf{ = 0}}\,\,{\bf{,Var}}\left( {\bf{\mu }} \right){\bf{ = 1,E}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{2}}}\,\,{\bf{and}}\,\,{\bf{Var}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}\)

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