$alpha$-Quantile for Gamma distribution $Gamma(n,frac{1}{2})$
1
$begingroup$
Let $XsimGamma(n,frac{1}{2})$ and $alphain(0,1)$. I know that
begin{align}
P(Xle s)&=int_0^ sfrac{1}{2^nGamma(n)}x^{n-1}e^{-frac{x}{2}}dx\
&=frac{1}{2(n-1)!}int_0^sBig(frac{x}{2}Big)^{n-1}e^{-frac{x}{2}}dx\
&=frac{1}{(n-1)!}int_0^{s/2} t^{n-1} e^{-t}dt\
end{align}
Now I wonder how can I find $u,v$ such that
$$frac{alpha}{2}ge frac{1}{(n-1)!}int_0^{u/2} t^{n-1} e^{-t}dt=P(Xle u),$$
$$1-frac{alpha}{2} le frac{1}{(n-1)!}int_0^{v/2} t^{n-1} e^{-t}dt= P(Xle v)text{ ?}$$
Maybe there is an easy argument... Thanks in advance for any help!
real-analysis probability analysis probability-theory statistics
asked Dec 18 '18 at 0:11
user408858user408858
482213
$endgroup$
add a comment |
1
$begingroup$
Let $XsimGamma(n,frac{1}{2})$ and $alphain(0,1)$. I know that
begin{align}
P(Xle s)&=int_0^ sfrac{1}{2^nGamma(n)}x^{n-1}e^{-frac{x}{2}}dx\
&=frac{1}{2(n-1)!}int_0^sBig(frac{x}{2}Big)^{n-1}e^{-frac{x}{2}}dx\
&=frac{1}{(n-1)!}int_0^{s/2} t^{n-1} e^{-t}dt\
end{align}
Now I wonder how can I find $u,v$ such that
$$frac{alpha}{2}ge frac{1}{(n-1)!}int_0^{u/2} t^{n-1} e^{-t}dt=P(Xle u),$$
$$1-frac{alpha}{2} le frac{1}{(n-1)!}int_0^{v/2} t^{n-1} e^{-t}dt= P(Xle v)text{ ?}$$
Maybe there is an easy argument... Thanks in advance for any help!
real-analysis probability analysis probability-theory statistics
asked Dec 18 '18 at 0:11
user408858user408858
482213
$endgroup$
$begingroup$
Of course, the integral in the end is the same as in the definition of the Gammafunction.
$endgroup$
– user408858
Dec 18 '18 at 0:12
$begingroup$
If I understand your notation and if you need the answers instead of hints how to do the calculus, then here is R code that will give you the answer for $n = 1, 2, dots, 50$ and $alpha = 0.05:$n = 1:50; L = qgamma(.025, n, 2); U = qgamma(.975, n ,2); cbind(n, L, U). For $n = 1,$ you have the exponential distribution, for which the calculus is trivial. For $n > 200,$ you should get good answers with normal approximation.
$endgroup$
– BruceET
Dec 18 '18 at 0:28
$begingroup$
Oh! Do I get you right, that there are these tables in backmatter of statistic books for exactly these kind of questions? :-)
$endgroup$
– user408858
Dec 18 '18 at 0:30
$begingroup$
You might be able to use widely available printed tables of chi-squared distributions (with appriat modifications; halve DF to get gamma shape). Some printed tables have only the upper tail. Look around. If you're allergic to R, I suppose MatLab and such will do much the same computations. // But R is free, devotedly bug free for basic stuff like this, and trivial to install. See Wikipedia. // Codeqchisq(.95, 4)andqgamma(.95, 2, 1/2)both return 9.487729
$endgroup$
– BruceET
Dec 18 '18 at 1:08
$begingroup$
Actually one can just take the $1-frac{alpha}{2}$-quantile by definition(?) or just calculate it like @BruceET did. :-)
$endgroup$
– user408858
Dec 20 '18 at 0:46
add a comment |
1
1
1
$begingroup$
Let $XsimGamma(n,frac{1}{2})$ and $alphain(0,1)$. I know that
begin{align}
P(Xle s)&=int_0^ sfrac{1}{2^nGamma(n)}x^{n-1}e^{-frac{x}{2}}dx\
&=frac{1}{2(n-1)!}int_0^sBig(frac{x}{2}Big)^{n-1}e^{-frac{x}{2}}dx\
&=frac{1}{(n-1)!}int_0^{s/2} t^{n-1} e^{-t}dt\
end{align}
Now I wonder how can I find $u,v$ such that
$$frac{alpha}{2}ge frac{1}{(n-1)!}int_0^{u/2} t^{n-1} e^{-t}dt=P(Xle u),$$
$$1-frac{alpha}{2} le frac{1}{(n-1)!}int_0^{v/2} t^{n-1} e^{-t}dt= P(Xle v)text{ ?}$$
Maybe there is an easy argument... Thanks in advance for any help!
real-analysis probability analysis probability-theory statistics
asked Dec 18 '18 at 0:11
user408858user408858
482213
$endgroup$
Let $XsimGamma(n,frac{1}{2})$ and $alphain(0,1)$. I know that
begin{align}
P(Xle s)&=int_0^ sfrac{1}{2^nGamma(n)}x^{n-1}e^{-frac{x}{2}}dx\
&=frac{1}{2(n-1)!}int_0^sBig(frac{x}{2}Big)^{n-1}e^{-frac{x}{2}}dx\
&=frac{1}{(n-1)!}int_0^{s/2} t^{n-1} e^{-t}dt\
end{align}
Now I wonder how can I find $u,v$ such that
$$frac{alpha}{2}ge frac{1}{(n-1)!}int_0^{u/2} t^{n-1} e^{-t}dt=P(Xle u),$$
$$1-frac{alpha}{2} le frac{1}{(n-1)!}int_0^{v/2} t^{n-1} e^{-t}dt= P(Xle v)text{ ?}$$
Maybe there is an easy argument... Thanks in advance for any help!
real-analysis probability analysis probability-theory statistics
real-analysis probability analysis probability-theory statistics
asked Dec 18 '18 at 0:11
user408858user408858
482213
asked Dec 18 '18 at 0:11
user408858user408858
482213
edited Dec 18 '18 at 0:21
user408858
asked Dec 18 '18 at 0:11
user408858user408858
482213
asked Dec 18 '18 at 0:11
user408858user408858
482213
asked Dec 18 '18 at 0:11
user408858user408858
482213
482213
$begingroup$
Of course, the integral in the end is the same as in the definition of the Gammafunction.
$endgroup$
– user408858
Dec 18 '18 at 0:12
$begingroup$
If I understand your notation and if you need the answers instead of hints how to do the calculus, then here is R code that will give you the answer for $n = 1, 2, dots, 50$ and $alpha = 0.05:$n = 1:50; L = qgamma(.025, n, 2); U = qgamma(.975, n ,2); cbind(n, L, U). For $n = 1,$ you have the exponential distribution, for which the calculus is trivial. For $n > 200,$ you should get good answers with normal approximation.
$endgroup$
– BruceET
Dec 18 '18 at 0:28
$begingroup$
Oh! Do I get you right, that there are these tables in backmatter of statistic books for exactly these kind of questions? :-)
$endgroup$
– user408858
Dec 18 '18 at 0:30
$begingroup$
You might be able to use widely available printed tables of chi-squared distributions (with appriat modifications; halve DF to get gamma shape). Some printed tables have only the upper tail. Look around. If you're allergic to R, I suppose MatLab and such will do much the same computations. // But R is free, devotedly bug free for basic stuff like this, and trivial to install. See Wikipedia. // Codeqchisq(.95, 4)andqgamma(.95, 2, 1/2)both return 9.487729
$endgroup$
– BruceET
Dec 18 '18 at 1:08
$begingroup$
Actually one can just take the $1-frac{alpha}{2}$-quantile by definition(?) or just calculate it like @BruceET did. :-)
$endgroup$
– user408858
Dec 20 '18 at 0:46
add a comment |
$begingroup$
Of course, the integral in the end is the same as in the definition of the Gammafunction.
$endgroup$
– user408858
Dec 18 '18 at 0:12
$begingroup$
If I understand your notation and if you need the answers instead of hints how to do the calculus, then here is R code that will give you the answer for $n = 1, 2, dots, 50$ and $alpha = 0.05:$n = 1:50; L = qgamma(.025, n, 2); U = qgamma(.975, n ,2); cbind(n, L, U). For $n = 1,$ you have the exponential distribution, for which the calculus is trivial. For $n > 200,$ you should get good answers with normal approximation.
$endgroup$
– BruceET
Dec 18 '18 at 0:28
$begingroup$
Oh! Do I get you right, that there are these tables in backmatter of statistic books for exactly these kind of questions? :-)
$endgroup$
– user408858
Dec 18 '18 at 0:30
$begingroup$
You might be able to use widely available printed tables of chi-squared distributions (with appriat modifications; halve DF to get gamma shape). Some printed tables have only the upper tail. Look around. If you're allergic to R, I suppose MatLab and such will do much the same computations. // But R is free, devotedly bug free for basic stuff like this, and trivial to install. See Wikipedia. // Codeqchisq(.95, 4)andqgamma(.95, 2, 1/2)both return 9.487729
$endgroup$
– BruceET
Dec 18 '18 at 1:08
$begingroup$
Actually one can just take the $1-frac{alpha}{2}$-quantile by definition(?) or just calculate it like @BruceET did. :-)
$endgroup$
– user408858
Dec 20 '18 at 0:46
$begingroup$
Of course, the integral in the end is the same as in the definition of the Gammafunction.
$endgroup$
– user408858
Dec 18 '18 at 0:12
$begingroup$
Of course, the integral in the end is the same as in the definition of the Gammafunction.
$endgroup$
– user408858
Dec 18 '18 at 0:12
$begingroup$
If I understand your notation and if you need the answers instead of hints how to do the calculus, then here is R code that will give you the answer for $n = 1, 2, dots, 50$ and $alpha = 0.05:$
n = 1:50; L = qgamma(.025, n, 2); U = qgamma(.975, n ,2); cbind(n, L, U). For $n = 1,$ you have the exponential distribution, for which the calculus is trivial. For $n > 200,$ you should get good answers with normal approximation.$endgroup$
– BruceET
Dec 18 '18 at 0:28
$begingroup$
If I understand your notation and if you need the answers instead of hints how to do the calculus, then here is R code that will give you the answer for $n = 1, 2, dots, 50$ and $alpha = 0.05:$
n = 1:50; L = qgamma(.025, n, 2); U = qgamma(.975, n ,2); cbind(n, L, U). For $n = 1,$ you have the exponential distribution, for which the calculus is trivial. For $n > 200,$ you should get good answers with normal approximation.$endgroup$
– BruceET
Dec 18 '18 at 0:28
$begingroup$
Oh! Do I get you right, that there are these tables in backmatter of statistic books for exactly these kind of questions? :-)
$endgroup$
– user408858
Dec 18 '18 at 0:30
$begingroup$
Oh! Do I get you right, that there are these tables in backmatter of statistic books for exactly these kind of questions? :-)
$endgroup$
– user408858
Dec 18 '18 at 0:30
$begingroup$
You might be able to use widely available printed tables of chi-squared distributions (with appriat modifications; halve DF to get gamma shape). Some printed tables have only the upper tail. Look around. If you're allergic to R, I suppose MatLab and such will do much the same computations. // But R is free, devotedly bug free for basic stuff like this, and trivial to install. See Wikipedia. // Code
qchisq(.95, 4) and qgamma(.95, 2, 1/2) both return 9.487729$endgroup$
– BruceET
Dec 18 '18 at 1:08
$begingroup$
You might be able to use widely available printed tables of chi-squared distributions (with appriat modifications; halve DF to get gamma shape). Some printed tables have only the upper tail. Look around. If you're allergic to R, I suppose MatLab and such will do much the same computations. // But R is free, devotedly bug free for basic stuff like this, and trivial to install. See Wikipedia. // Code
qchisq(.95, 4) and qgamma(.95, 2, 1/2) both return 9.487729$endgroup$
– BruceET
Dec 18 '18 at 1:08
$begingroup$
Actually one can just take the $1-frac{alpha}{2}$-quantile by definition(?) or just calculate it like @BruceET did. :-)
$endgroup$
– user408858
Dec 20 '18 at 0:46
$begingroup$
Actually one can just take the $1-frac{alpha}{2}$-quantile by definition(?) or just calculate it like @BruceET did. :-)
$endgroup$
– user408858
Dec 20 '18 at 0:46
add a comment |
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$begingroup$
Of course, the integral in the end is the same as in the definition of the Gammafunction.
$endgroup$
– user408858
Dec 18 '18 at 0:12
$begingroup$
If I understand your notation and if you need the answers instead of hints how to do the calculus, then here is R code that will give you the answer for $n = 1, 2, dots, 50$ and $alpha = 0.05:$
n = 1:50; L = qgamma(.025, n, 2); U = qgamma(.975, n ,2); cbind(n, L, U). For $n = 1,$ you have the exponential distribution, for which the calculus is trivial. For $n > 200,$ you should get good answers with normal approximation.$endgroup$
– BruceET
Dec 18 '18 at 0:28
$begingroup$
Oh! Do I get you right, that there are these tables in backmatter of statistic books for exactly these kind of questions? :-)
$endgroup$
– user408858
Dec 18 '18 at 0:30
$begingroup$
You might be able to use widely available printed tables of chi-squared distributions (with appriat modifications; halve DF to get gamma shape). Some printed tables have only the upper tail. Look around. If you're allergic to R, I suppose MatLab and such will do much the same computations. // But R is free, devotedly bug free for basic stuff like this, and trivial to install. See Wikipedia. // Code
qchisq(.95, 4)andqgamma(.95, 2, 1/2)both return 9.487729$endgroup$
– BruceET
Dec 18 '18 at 1:08
$begingroup$
Actually one can just take the $1-frac{alpha}{2}$-quantile by definition(?) or just calculate it like @BruceET did. :-)
$endgroup$
– user408858
Dec 20 '18 at 0:46