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Chapter 8: 8.2 - Q2 E (page 472)

Find the mode of the蠂2 distribution withmdegrees offreedom(m=1,2, . . .).

Short Answer

Expert verified

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

Step by step solution

01

Given information

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

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 situation described in Example 8.2.3. How small would蟽2 need to be in order for Pr(Y鈮�0.09)鈮�0.9?

The study on acid concentration in cheese included a total of 30 lactic acid measurements, the 10 given in Example 8.5.4 on page 487 and the following additional 20:

1.68, 1.9, 1.06, 1.3, 1.52, 1.74, 1.16, 1.49, 1.63, 1.99, 1.15, 1.33, 1.44, 2.01, 1.31, 1.46, 1.72, 1.25, 1.08, 1.25.

a. Using the same prior as in Example 8.6.2 on page 498, compute the posterior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\) based on all 30 observations.

b. Use the posterior distribution found in Example 8.6.2 on page 498 as if it were the prior distribution before observing the 20 observations listed in this problem. Use these 20 new observations to find the posterior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\)and compare the result to the answer to part (a).

Suppose that \({X_1},...,{X_n}\) form a random sample from the normal distribution with mean 渭 and variance \({\sigma ^2}\) . Assuming that the sample size n is 16, determine the values of the following probabilities:

\(\begin{align}a.\,\,\,\,P\left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \mu } \right)}^2} \le 2{\sigma ^2}} } \right)\\b.\,\,\,\,P\left( {\frac{1}{2}{\sigma ^2} \le \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2} \le 2{\sigma ^2}} } \right)\end{align}\)

Suppose that\({X_1}...{X_n}\)form a random sample from the normal distribution with mean 0 and unknown standard deviation\(\sigma > 0\). Find the lower bound specified by the information inequality for the variance of any unbiased estimator of\(\log \sigma \).

Determine whether or not each of the five following matrices is orthogonal:

  1. \(\left( {\begin{align}{\bf{0}}&{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}\end{align}} \right)\)
  2. \(\left( {\begin{align}{{\bf{0}}{\bf{.8}}}&{\bf{0}}&{{\bf{0}}{\bf{.6}}}\\{{\bf{ - 0}}{\bf{.6}}}&{\bf{0}}&{{\bf{0}}{\bf{.8}}}\\{\bf{0}}&{{\bf{ - 1}}}&{\bf{0}}\end{align}} \right)\)
  3. \(\left( {\begin{align}{{\bf{0}}{\bf{.8}}}&{\bf{0}}&{{\bf{0}}{\bf{.6}}}\\{{\bf{ - 0}}{\bf{.6}}}&{\bf{0}}&{{\bf{0}}{\bf{.8}}}\\{\bf{0}}&{{\bf{0}}{\bf{.5}}}&{\bf{0}}\end{align}} \right)\)
  4. \(\left( {\begin{align}{}{{\bf{ - }}\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}\\{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{{\bf{ - }}\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}\\{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{{\bf{ - }}\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}\end{align}} \right)\)
  5. \(\left( {\begin{align}{}{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}\\{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}\\{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}\\{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}\end{align}} \right)\)
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