Mixing
On a Bayesian hypothesis testing
$begingroup$
Let $X_1,...,X_n$ be a random sample and $lambda >0$ be a parameter, with $X_i |lambda sim Poisson (lambda)$ and $lambda sim Gamma(alpha, beta) (lambda)=dfrac {1}{Gamma(alpha) beta^alpha} lambda ^{alpha-1} e^{-lambda / beta}$.
I want to find a Bayesian test of hypothesis for $H_0 : lambda le lambda_0$ against $H_1: lambda > lambda_1$ .
Now I know that a simple rule is that accept $H_0 :$ If $P(lambda le lambda_0|Y=y) ge P(lambda > lambda_0 | Y=y) $ .
Now $Y=sum_{i=1}^n X_i$ is a sufficient statistics , and the posterior pdf of $lambda |Y=y$ is a constant (depending on $y$ but independent of $lambda$ ) times $lambda ^{y+alpha -1} e^{-nlambda -frac {lambda}{ beta}}$ , so $lambda|Y=y sim Gamma (y+alpha, dfrac {beta}{nbeta+1}), so $ $P(lambda le lambda_0|Y=y) ge P(lambda > lambda_0 | Y=y)$ is equivalent to saying
$dfrac {1}{Gamma (y+alpha ) } gamma (y+alpha,nlambda_0 +frac {lambda_0}{ beta})ge 1-dfrac {1}{Gamma (y+alpha ) } gamma (y+alpha,nlambda_0 +frac {lambda_0}{ beta})$, so
$lambda_0 ge $ median of $Gamma (y+alpha, dfrac {beta}{nbeta+1}) $ .
But I don't know what to do further. Am I even on the right track ?
Please help .
probability-theory statistics probability-distributions bayesian hypothesis-testing
asked Jan 13 at 5:24
user521337user521337
1,1801416
$endgroup$
add a comment |
$begingroup$
Let $X_1,...,X_n$ be a random sample and $lambda >0$ be a parameter, with $X_i |lambda sim Poisson (lambda)$ and $lambda sim Gamma(alpha, beta) (lambda)=dfrac {1}{Gamma(alpha) beta^alpha} lambda ^{alpha-1} e^{-lambda / beta}$.
I want to find a Bayesian test of hypothesis for $H_0 : lambda le lambda_0$ against $H_1: lambda > lambda_1$ .
Now I know that a simple rule is that accept $H_0 :$ If $P(lambda le lambda_0|Y=y) ge P(lambda > lambda_0 | Y=y) $ .
Now $Y=sum_{i=1}^n X_i$ is a sufficient statistics , and the posterior pdf of $lambda |Y=y$ is a constant (depending on $y$ but independent of $lambda$ ) times $lambda ^{y+alpha -1} e^{-nlambda -frac {lambda}{ beta}}$ , so $lambda|Y=y sim Gamma (y+alpha, dfrac {beta}{nbeta+1}), so $ $P(lambda le lambda_0|Y=y) ge P(lambda > lambda_0 | Y=y)$ is equivalent to saying
$dfrac {1}{Gamma (y+alpha ) } gamma (y+alpha,nlambda_0 +frac {lambda_0}{ beta})ge 1-dfrac {1}{Gamma (y+alpha ) } gamma (y+alpha,nlambda_0 +frac {lambda_0}{ beta})$, so
$lambda_0 ge $ median of $Gamma (y+alpha, dfrac {beta}{nbeta+1}) $ .
But I don't know what to do further. Am I even on the right track ?
Please help .
probability-theory statistics probability-distributions bayesian hypothesis-testing
asked Jan 13 at 5:24
user521337user521337
1,1801416
$endgroup$
$begingroup$
Yes, the rule you chose is equivalent to deciding based of which side of $lambda_0$ the median of the posterior is on, and it seems like you've computed the posterior correctly. Assuming this is the rule you want, I don't know that there's anything you can do further until it's actually time to put some numbers in. There's no closed form expression for that median.
$endgroup$
– spaceisdarkgreen
Jan 13 at 6:42
add a comment |
$begingroup$
Let $X_1,...,X_n$ be a random sample and $lambda >0$ be a parameter, with $X_i |lambda sim Poisson (lambda)$ and $lambda sim Gamma(alpha, beta) (lambda)=dfrac {1}{Gamma(alpha) beta^alpha} lambda ^{alpha-1} e^{-lambda / beta}$.
I want to find a Bayesian test of hypothesis for $H_0 : lambda le lambda_0$ against $H_1: lambda > lambda_1$ .
Now I know that a simple rule is that accept $H_0 :$ If $P(lambda le lambda_0|Y=y) ge P(lambda > lambda_0 | Y=y) $ .
Now $Y=sum_{i=1}^n X_i$ is a sufficient statistics , and the posterior pdf of $lambda |Y=y$ is a constant (depending on $y$ but independent of $lambda$ ) times $lambda ^{y+alpha -1} e^{-nlambda -frac {lambda}{ beta}}$ , so $lambda|Y=y sim Gamma (y+alpha, dfrac {beta}{nbeta+1}), so $ $P(lambda le lambda_0|Y=y) ge P(lambda > lambda_0 | Y=y)$ is equivalent to saying
$dfrac {1}{Gamma (y+alpha ) } gamma (y+alpha,nlambda_0 +frac {lambda_0}{ beta})ge 1-dfrac {1}{Gamma (y+alpha ) } gamma (y+alpha,nlambda_0 +frac {lambda_0}{ beta})$, so
$lambda_0 ge $ median of $Gamma (y+alpha, dfrac {beta}{nbeta+1}) $ .
But I don't know what to do further. Am I even on the right track ?
Please help .
probability-theory statistics probability-distributions bayesian hypothesis-testing
asked Jan 13 at 5:24
user521337user521337
1,1801416
$endgroup$
Let $X_1,...,X_n$ be a random sample and $lambda >0$ be a parameter, with $X_i |lambda sim Poisson (lambda)$ and $lambda sim Gamma(alpha, beta) (lambda)=dfrac {1}{Gamma(alpha) beta^alpha} lambda ^{alpha-1} e^{-lambda / beta}$.
I want to find a Bayesian test of hypothesis for $H_0 : lambda le lambda_0$ against $H_1: lambda > lambda_1$ .
Now I know that a simple rule is that accept $H_0 :$ If $P(lambda le lambda_0|Y=y) ge P(lambda > lambda_0 | Y=y) $ .
Now $Y=sum_{i=1}^n X_i$ is a sufficient statistics , and the posterior pdf of $lambda |Y=y$ is a constant (depending on $y$ but independent of $lambda$ ) times $lambda ^{y+alpha -1} e^{-nlambda -frac {lambda}{ beta}}$ , so $lambda|Y=y sim Gamma (y+alpha, dfrac {beta}{nbeta+1}), so $ $P(lambda le lambda_0|Y=y) ge P(lambda > lambda_0 | Y=y)$ is equivalent to saying
$dfrac {1}{Gamma (y+alpha ) } gamma (y+alpha,nlambda_0 +frac {lambda_0}{ beta})ge 1-dfrac {1}{Gamma (y+alpha ) } gamma (y+alpha,nlambda_0 +frac {lambda_0}{ beta})$, so
$lambda_0 ge $ median of $Gamma (y+alpha, dfrac {beta}{nbeta+1}) $ .
But I don't know what to do further. Am I even on the right track ?
Please help .
probability-theory statistics probability-distributions bayesian hypothesis-testing
probability-theory statistics probability-distributions bayesian hypothesis-testing
asked Jan 13 at 5:24
user521337user521337
1,1801416
asked Jan 13 at 5:24
user521337user521337
1,1801416
asked Jan 13 at 5:24
user521337user521337
1,1801416
asked Jan 13 at 5:24
user521337user521337
1,1801416
asked Jan 13 at 5:24
user521337user521337
1,1801416
1,1801416
$begingroup$
Yes, the rule you chose is equivalent to deciding based of which side of $lambda_0$ the median of the posterior is on, and it seems like you've computed the posterior correctly. Assuming this is the rule you want, I don't know that there's anything you can do further until it's actually time to put some numbers in. There's no closed form expression for that median.
$endgroup$
– spaceisdarkgreen
Jan 13 at 6:42
add a comment |
$begingroup$
Yes, the rule you chose is equivalent to deciding based of which side of $lambda_0$ the median of the posterior is on, and it seems like you've computed the posterior correctly. Assuming this is the rule you want, I don't know that there's anything you can do further until it's actually time to put some numbers in. There's no closed form expression for that median.
$endgroup$
– spaceisdarkgreen
Jan 13 at 6:42
$begingroup$
Yes, the rule you chose is equivalent to deciding based of which side of $lambda_0$ the median of the posterior is on, and it seems like you've computed the posterior correctly. Assuming this is the rule you want, I don't know that there's anything you can do further until it's actually time to put some numbers in. There's no closed form expression for that median.
$endgroup$
– spaceisdarkgreen
Jan 13 at 6:42
$begingroup$
Yes, the rule you chose is equivalent to deciding based of which side of $lambda_0$ the median of the posterior is on, and it seems like you've computed the posterior correctly. Assuming this is the rule you want, I don't know that there's anything you can do further until it's actually time to put some numbers in. There's no closed form expression for that median.
$endgroup$
– spaceisdarkgreen
Jan 13 at 6:42
add a comment |
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$begingroup$
Yes, the rule you chose is equivalent to deciding based of which side of $lambda_0$ the median of the posterior is on, and it seems like you've computed the posterior correctly. Assuming this is the rule you want, I don't know that there's anything you can do further until it's actually time to put some numbers in. There's no closed form expression for that median.
$endgroup$
– spaceisdarkgreen
Jan 13 at 6:42