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| + | =Question= | ||
<math> | <math> | ||
H_0:\ \mbox{X has PDF}\ f_{X|\theta}(x|\theta_0)= | H_0:\ \mbox{X has PDF}\ f_{X|\theta}(x|\theta_0)= | ||
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</math> | </math> | ||
| − | + | Find the ML rule. | |
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| + | =Answer= | ||
To find the ML Rule we say pick <math>H_1\!</math> if <math>f_{X|\theta}(x|\theta_1)>f_{X|\theta}(x|\theta_0)\!</math> | To find the ML Rule we say pick <math>H_1\!</math> if <math>f_{X|\theta}(x|\theta_1)>f_{X|\theta}(x|\theta_0)\!</math> | ||
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<math> Pr[x\ge1/2|\theta=\theta_1] = \int_{1/2}^{1}1\, dx=1/2\! </math><br><br> | <math> Pr[x\ge1/2|\theta=\theta_1] = \int_{1/2}^{1}1\, dx=1/2\! </math><br><br> | ||
| + | ---- | ||
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| + | [[Main_Page_ECE302Fall2008sanghavi|Back to ECE302 Fall 2008 Prof. Sanghavi]] | ||
Revision as of 12:46, 22 November 2011
Question
$ H_0:\ \mbox{X has PDF}\ f_{X|\theta}(x|\theta_0)= \begin{cases} 2x & \mbox{for }0 \le x \le 1 \\ 0 & \mbox{else} \end{cases} $
$ H_1:\ \mbox{X has PDF}\ f_{X|\theta}(x|\theta_1)= \begin{cases} 1 & \mbox{for }0 \le x \le 1 \\ 0 & \mbox{else} \end{cases} $
Find the ML rule.
Answer
To find the ML Rule we say pick $ H_1\! $ if $ f_{X|\theta}(x|\theta_1)>f_{X|\theta}(x|\theta_0)\! $
Or in otherwords pick $ H_1\! $ if $ 1>2x\! $ Thus,
$ \mbox{ML Rule: } \begin{cases} \mbox{say }H_1 &\mbox{if }x<1/2 \\ \mbox{say }H_0 &\mbox{else} \end{cases} $
Type I Error: A Type I error is the $ Pr[\mbox{say } H_1|H_0]\! $ this is equivalent to saying $ Pr[x<1/2|\theta=\theta_0]\! $ we calculate this using integration
$ Pr[x<1/2|\theta=\theta_0] = \int_{0}^{1/2}2x\, dx=1/4\! $
Type II Error:A Type II error is the $ Pr[\mbox{say } H_0|H_1]\! $ this is equivalent to saying $ Pr[x\ge1/2|\theta=\theta_1] $ we calculate this using integration
$ Pr[x\ge1/2|\theta=\theta_1] = \int_{1/2}^{1}1\, dx=1/2\! $
