1. ## Confidence Interval

Suppose that $Y$ is normally distributed with mean 0 and unknown variance $\sigma ^2$. Then $\frac{Y^2}{\sigma^2}$ has a $\chi^2$ with 1 df. Use the pivotal quantity $\frac{Y^2}{\sigma^2}$ to find:

a) 95% confidence interval for $\sigma^2$
b) 95% upper confidence limit of $\sigma^2$

Solution:
I based my work on http://www.mathhelpforum.com/math-he...-interval.html which gave me

$f(y) = \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}}$

then $P(Y<\alpha) = \int^{\alpha}_0 \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.025$

$P(Y<\alpha) = \frac{1}{\sqrt{2\pi}} \int^{\alpha}_0 y^{(1/2)-1}e^{-y/2} \ dy = 0.025$

if I make the substitution $y=2x$ and $\frac{dy}{dx} =2 \longrightarrow dy = 2dx$ then I get:

$P(2X<\alpha) = \frac{1}{\sqrt{2\pi}} \int^{\alpha}_0 (2x)^{(1/2)-1}e^{-x} \ 2dx = 0.025$

$P(2X<\alpha) = \frac{1}{\sqrt{\pi}} \int^{\alpha}_0 x^{(1/2)-1}e^{-x} \ dx = 0.025$

now this looks a lot like a Gamma function, but my upper bound is $\alpha$ and not $\infty$, which is throwing me off.

$P(Y>\beta) = \int^{\infty}_{\beta} \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.025$

$P(Y>\beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} y^{(1/2)-1}e^{-y/2} \ dy = 0.025$

same substitution as before

$P(2X>\beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} (2x)^{(1/2)-1}e^{-x} \ 2dx = 0.025$

$P(2X>\beta) = \frac{1}{\sqrt{\pi}} \int^{\infty}_{\beta} x^{(1/2)-1}e^{-x} \ dx = 0.025$

again this is looking a lot like a Gamma function, but this this my lower bound is not 0 it's $\beta$

the answer in the back of the book is $\left(\frac{Y^2}{5.02389},\ \frac{Y^2}{0.71072} \right)
$

b) $P(Y \geq \beta) = \int^{\infty}_{\beta} \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.95$

$P(Y \geq \beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} y^{(1/2)-1}e^{-y/2} \ dy = 0.95$

same substitution and same argument as before. I get:

$P(2X \geq \beta) = \frac{1}{\sqrt{\pi}} \int^{\infty}_{\beta} x^{(1/2)-1}e^{-x} \ dx = 0.95$

with the solution in the back of the book being $\frac{Y^2}{0.0039321}$

2. Originally Posted by lllll
Suppose that $Y$ is normally distributed with mean 0 and unknown variance $\sigma ^2$. Then $\frac{Y^2}{\sigma^2}$ has a $\chi^2$ with 1 df. Use the pivotal quantity $\frac{Y^2}{\sigma^2}$ to find:

a) 95% confidence interval for $\sigma^2$
b) 95% upper confidence limit of $\sigma^2$

Solution:
I based my work on http://www.mathhelpforum.com/math-he...-interval.html which gave me

$f(y) = \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}}$

Mr F says: All corrections are given in red:

then $P({\color{red}U}<\alpha) = \int^{\alpha}_0 \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.025$ Mr F says: No. The random variable here is not Y. It's ${\color{red}\frac{Y^2}{\sigma^2} = U}$, say.

$P( {\color{red}U} <\alpha) = \frac{1}{\sqrt{2\pi}} \int^{\alpha}_0 y^{(1/2)-1}e^{-y/2} \ dy = 0.025$

if I make the substitution $y=2x$ and $\frac{dy}{dx} =2 \longrightarrow dy = 2dx$ then I get:

$P({\color{red}U}<\alpha) = \frac{1}{\sqrt{2\pi}} \int^{\alpha{\color{red}/2}}_0 (2x)^{(1/2)-1}e^{-x} \ 2dx = 0.025$ Mr F says: The substitution is a change of variable only in the integral, NOT a change of random variable! That 2X you had should still be a U. Also, if you introduce a change of variable, you have to change the integral terminals too!

Actually, my next comment (below) shows why there's actually no point in making a substitution.

$P({\color{red}U}<\alpha) = \frac{1}{\sqrt{\pi}} \int^{\alpha{\color{red}/2}}_0 x^{(1/2)-1}e^{-x} \ dx = 0.025$

now this looks a lot like a Gamma function, but my upper bound is $\alpha$ and not $\infty$, which is throwing me off.

Mr F says: This equation can only be readily solved for ${\color{red}\alpha}$ by using technology. In fact, I'm not sure it can be solved at all without using technology.

Everything I've said above applies below too.

$P(Y>\beta) = \int^{\infty}_{\beta} \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.025$

$P(Y>\beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} y^{(1/2)-1}e^{-y/2} \ dy = 0.025$

same substitution as before

$P(2X>\beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} (2x)^{(1/2)-1}e^{-x} \ 2dx = 0.025$

$P(2X>\beta) = \frac{1}{\sqrt{\pi}} \int^{\infty}_{\beta} x^{(1/2)-1}e^{-x} \ dx = 0.025$

again this is looking a lot like a Gamma function, but this this my lower bound is not 0 it's $\beta$

the answer in the back of the book is $\left(\frac{Y^2}{5.02389},\ \frac{Y^2}{0.71072} \right)
$

b) $P(Y \geq \beta) = \int^{\infty}_{\beta} \frac{y^{(1/2)-1}e^{-y/2}}{2^{1/2}\Gamma{(1/2)}} \ dy = 0.95$

$P(Y \geq \beta) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{\beta} y^{(1/2)-1}e^{-y/2} \ dy = 0.95$

same substitution and same argument as before. I get:

$P(2X \geq \beta) = \frac{1}{\sqrt{\pi}} \int^{\infty}_{\beta} x^{(1/2)-1}e^{-x} \ dx = 0.95$

with the solution in the back of the book being $\frac{Y^2}{0.0039321}$
..