Beta distribution: find the parameter $alpha$ of $mathcal{B}e(alpha,frac{1}{3})$
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I have this variable with beta distribution : $Y sim mathcal{B}e(alpha,frac{1}{3})$.
I have to find the value of $alpha$ such as :
$P(Y leq 0.416) =0.2 $
Formally for $alpha geq 0$ , $beta geq 0$ and $0 leq y leq 1$ the CDF function of $Y$ at 0.416 is:
$P(Y leq 0.416) = frac{Gamma(alpha + beta)}{Gamma(alpha)Gamma(beta)} int_0^{0.416} t^{alpha-1} (1-t)^{beta-1} dt=0.2$
I am not sure how to proceed. Thanks for the help in advance!!
probability integration probability-distributions gamma-function beta-function
asked Dec 13 '18 at 15:04
andrewandrew
698
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add a comment |
$begingroup$
I have this variable with beta distribution : $Y sim mathcal{B}e(alpha,frac{1}{3})$.
I have to find the value of $alpha$ such as :
$P(Y leq 0.416) =0.2 $
Formally for $alpha geq 0$ , $beta geq 0$ and $0 leq y leq 1$ the CDF function of $Y$ at 0.416 is:
$P(Y leq 0.416) = frac{Gamma(alpha + beta)}{Gamma(alpha)Gamma(beta)} int_0^{0.416} t^{alpha-1} (1-t)^{beta-1} dt=0.2$
I am not sure how to proceed. Thanks for the help in advance!!
probability integration probability-distributions gamma-function beta-function
asked Dec 13 '18 at 15:04
andrewandrew
698
$endgroup$
add a comment |
$begingroup$
I have this variable with beta distribution : $Y sim mathcal{B}e(alpha,frac{1}{3})$.
I have to find the value of $alpha$ such as :
$P(Y leq 0.416) =0.2 $
Formally for $alpha geq 0$ , $beta geq 0$ and $0 leq y leq 1$ the CDF function of $Y$ at 0.416 is:
$P(Y leq 0.416) = frac{Gamma(alpha + beta)}{Gamma(alpha)Gamma(beta)} int_0^{0.416} t^{alpha-1} (1-t)^{beta-1} dt=0.2$
I am not sure how to proceed. Thanks for the help in advance!!
probability integration probability-distributions gamma-function beta-function
asked Dec 13 '18 at 15:04
andrewandrew
698
$endgroup$
I have this variable with beta distribution : $Y sim mathcal{B}e(alpha,frac{1}{3})$.
I have to find the value of $alpha$ such as :
$P(Y leq 0.416) =0.2 $
Formally for $alpha geq 0$ , $beta geq 0$ and $0 leq y leq 1$ the CDF function of $Y$ at 0.416 is:
$P(Y leq 0.416) = frac{Gamma(alpha + beta)}{Gamma(alpha)Gamma(beta)} int_0^{0.416} t^{alpha-1} (1-t)^{beta-1} dt=0.2$
I am not sure how to proceed. Thanks for the help in advance!!
probability integration probability-distributions gamma-function beta-function
probability integration probability-distributions gamma-function beta-function
asked Dec 13 '18 at 15:04
andrewandrew
698
asked Dec 13 '18 at 15:04
andrewandrew
698
asked Dec 13 '18 at 15:04
andrewandrew
698
asked Dec 13 '18 at 15:04
andrewandrew
698
asked Dec 13 '18 at 15:04
andrewandrew
698
698
add a comment |
add a comment |
1 Answer
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$begingroup$
Maple gives the CDF for $mathcal{B}e(alpha,beta)$ as
$$ {frac {Gamma left( alpha+beta right) {y}^{alpha}
{mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}}{Gamma left(
alpha right) Gamma left( beta right) alpha}}
$$
In any case, you need to use numerical methods to solve $F(0.416) = 0.2$.
Maple says $alpha = 0.8563203833$.
answered Dec 13 '18 at 15:12
Robert IsraelRobert Israel
325k23214468
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$begingroup$
The question is stated as though an analytical solution is expected...but I agree that numerical methods are the way to proceed here.
$endgroup$
– Math1000
Dec 13 '18 at 15:13
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thank you very much for the kind answer. But i don't understand the meaning of this quantity: ${mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}$. And how did you obtain the value of $alpha$? Thank you in advance @Robert Israel
$endgroup$
– andrew
Dec 13 '18 at 15:25
$begingroup$
Can i get a solution with the software R?
$endgroup$
– andrew
Dec 13 '18 at 16:40
$begingroup$
${}_2F_1(a,b; c; z)$ is a Gaussian hypergeometric function. The value of $alpha$ was obtained using a numerical solver (fsolve in Maple).
$endgroup$
– Robert Israel
Dec 13 '18 at 18:55
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Maple gives the CDF for $mathcal{B}e(alpha,beta)$ as
$$ {frac {Gamma left( alpha+beta right) {y}^{alpha}
{mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}}{Gamma left(
alpha right) Gamma left( beta right) alpha}}
$$
In any case, you need to use numerical methods to solve $F(0.416) = 0.2$.
Maple says $alpha = 0.8563203833$.
answered Dec 13 '18 at 15:12
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
The question is stated as though an analytical solution is expected...but I agree that numerical methods are the way to proceed here.
$endgroup$
– Math1000
Dec 13 '18 at 15:13
$begingroup$
thank you very much for the kind answer. But i don't understand the meaning of this quantity: ${mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}$. And how did you obtain the value of $alpha$? Thank you in advance @Robert Israel
$endgroup$
– andrew
Dec 13 '18 at 15:25
$begingroup$
Can i get a solution with the software R?
$endgroup$
– andrew
Dec 13 '18 at 16:40
$begingroup$
${}_2F_1(a,b; c; z)$ is a Gaussian hypergeometric function. The value of $alpha$ was obtained using a numerical solver (fsolve in Maple).
$endgroup$
– Robert Israel
Dec 13 '18 at 18:55
add a comment |
$begingroup$
Maple gives the CDF for $mathcal{B}e(alpha,beta)$ as
$$ {frac {Gamma left( alpha+beta right) {y}^{alpha}
{mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}}{Gamma left(
alpha right) Gamma left( beta right) alpha}}
$$
In any case, you need to use numerical methods to solve $F(0.416) = 0.2$.
Maple says $alpha = 0.8563203833$.
answered Dec 13 '18 at 15:12
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
The question is stated as though an analytical solution is expected...but I agree that numerical methods are the way to proceed here.
$endgroup$
– Math1000
Dec 13 '18 at 15:13
$begingroup$
thank you very much for the kind answer. But i don't understand the meaning of this quantity: ${mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}$. And how did you obtain the value of $alpha$? Thank you in advance @Robert Israel
$endgroup$
– andrew
Dec 13 '18 at 15:25
$begingroup$
Can i get a solution with the software R?
$endgroup$
– andrew
Dec 13 '18 at 16:40
$begingroup$
${}_2F_1(a,b; c; z)$ is a Gaussian hypergeometric function. The value of $alpha$ was obtained using a numerical solver (fsolve in Maple).
$endgroup$
– Robert Israel
Dec 13 '18 at 18:55
add a comment |
$begingroup$
Maple gives the CDF for $mathcal{B}e(alpha,beta)$ as
$$ {frac {Gamma left( alpha+beta right) {y}^{alpha}
{mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}}{Gamma left(
alpha right) Gamma left( beta right) alpha}}
$$
In any case, you need to use numerical methods to solve $F(0.416) = 0.2$.
Maple says $alpha = 0.8563203833$.
answered Dec 13 '18 at 15:12
Robert IsraelRobert Israel
325k23214468
$endgroup$
Maple gives the CDF for $mathcal{B}e(alpha,beta)$ as
$$ {frac {Gamma left( alpha+beta right) {y}^{alpha}
{mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}}{Gamma left(
alpha right) Gamma left( beta right) alpha}}
$$
In any case, you need to use numerical methods to solve $F(0.416) = 0.2$.
Maple says $alpha = 0.8563203833$.
answered Dec 13 '18 at 15:12
Robert IsraelRobert Israel
325k23214468
answered Dec 13 '18 at 15:12
Robert IsraelRobert Israel
325k23214468
answered Dec 13 '18 at 15:12
Robert IsraelRobert Israel
325k23214468
answered Dec 13 '18 at 15:12
Robert IsraelRobert Israel
325k23214468
325k23214468
$begingroup$
The question is stated as though an analytical solution is expected...but I agree that numerical methods are the way to proceed here.
$endgroup$
– Math1000
Dec 13 '18 at 15:13
$begingroup$
thank you very much for the kind answer. But i don't understand the meaning of this quantity: ${mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}$. And how did you obtain the value of $alpha$? Thank you in advance @Robert Israel
$endgroup$
– andrew
Dec 13 '18 at 15:25
$begingroup$
Can i get a solution with the software R?
$endgroup$
– andrew
Dec 13 '18 at 16:40
$begingroup$
${}_2F_1(a,b; c; z)$ is a Gaussian hypergeometric function. The value of $alpha$ was obtained using a numerical solver (fsolve in Maple).
$endgroup$
– Robert Israel
Dec 13 '18 at 18:55
add a comment |
$begingroup$
The question is stated as though an analytical solution is expected...but I agree that numerical methods are the way to proceed here.
$endgroup$
– Math1000
Dec 13 '18 at 15:13
$begingroup$
thank you very much for the kind answer. But i don't understand the meaning of this quantity: ${mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}$. And how did you obtain the value of $alpha$? Thank you in advance @Robert Israel
$endgroup$
– andrew
Dec 13 '18 at 15:25
$begingroup$
Can i get a solution with the software R?
$endgroup$
– andrew
Dec 13 '18 at 16:40
$begingroup$
${}_2F_1(a,b; c; z)$ is a Gaussian hypergeometric function. The value of $alpha$ was obtained using a numerical solver (fsolve in Maple).
$endgroup$
– Robert Israel
Dec 13 '18 at 18:55
$begingroup$
The question is stated as though an analytical solution is expected...but I agree that numerical methods are the way to proceed here.
$endgroup$
– Math1000
Dec 13 '18 at 15:13
$begingroup$
The question is stated as though an analytical solution is expected...but I agree that numerical methods are the way to proceed here.
$endgroup$
– Math1000
Dec 13 '18 at 15:13
$begingroup$
thank you very much for the kind answer. But i don't understand the meaning of this quantity: ${mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}$. And how did you obtain the value of $alpha$? Thank you in advance @Robert Israel
$endgroup$
– andrew
Dec 13 '18 at 15:25
$begingroup$
thank you very much for the kind answer. But i don't understand the meaning of this quantity: ${mbox{$_2$F$_1$}(alpha,1-beta;,1+alpha;,y)}$. And how did you obtain the value of $alpha$? Thank you in advance @Robert Israel
$endgroup$
– andrew
Dec 13 '18 at 15:25
$begingroup$
Can i get a solution with the software R?
$endgroup$
– andrew
Dec 13 '18 at 16:40
$begingroup$
Can i get a solution with the software R?
$endgroup$
– andrew
Dec 13 '18 at 16:40
$begingroup$
${}_2F_1(a,b; c; z)$ is a Gaussian hypergeometric function. The value of $alpha$ was obtained using a numerical solver (fsolve in Maple).
$endgroup$
– Robert Israel
Dec 13 '18 at 18:55
$begingroup$
${}_2F_1(a,b; c; z)$ is a Gaussian hypergeometric function. The value of $alpha$ was obtained using a numerical solver (fsolve in Maple).
$endgroup$
– Robert Israel
Dec 13 '18 at 18:55
add a comment |
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