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prove that $Y_1 sim Bin(n_1,pi)$ and $Y_2 sim Bin(n_2,cpi)$ is an exponential family

Showing that a random sum of logarithmic mass functions has negative binomial distributionPoisson rate regression for grouped data: How to derive alpha and betaFinding the Correct Sample Size N to calculate the $SEM$ in a Relative A/B Test with $X sim B(p_1,n_1)$ and $Y sim B(p_2,n_2)$How to show that $E(X^k)=npE((Y + 1)^k-1)$ where $XsimmathrmBin(n,p)$ and $Y sim mathrmBin(n-1,p)$.Simplify $E(max(X_1+Y_1, X_2+Y_2))$ when $X_1, Y_1, X_2$, and $Y_2$ are exponentially distributedMoment generating function of exponential familyCounterproof for sufficient statistic as well as an example for minimal sufficient statisticFinding $mathbbP(Y_1 < Y_2)$ of two independent exponential r.v.sSuppose $X_1$ and $Y_1$, and $X_2$ and $Y_2$ have identical distributions too. What about $(X_1,X_2)$ and $(Y_1,Y_2)$?Exponential family distribution and sufficient statistic.

0

$begingroup$

A statistical model for a data set y is an exponential family , with canonical parameter vector $theta= (theta_1,theta_2,..theta_k)$ and canonical statistic $t(y) = (t_1(y),t_2(y),..t_k(y))$ if $f$ has structure

$$f(y;theta) = a(theta)h(y) exp(theta^T t(y)).$$

Now,consider two binomially distributed random variables $Y_1 sim Bin(n_1,pi)$ and $Y_2 sim Bin(n_2,cpi)$ prove that the canonical statistic $t(Y_1,Y_2) = (v,y)^T$ is $v = Y_1,u = Y_1 + Y_2$

$binomn_1y_1 binomn_2y_2 pi^y_1(1-pi)^n_1-y_1 (cpi)^y_2(1-cpi)^n_2-y_2 $

The only thing I can come up with is the form:

$binomn_1y_1 binomn_2y_2 (1-pi)^n_1 Big( dfracpi1-piBig)^y_1+y_2 Big(dfrac(1-pi)c1-cpiBig)^y_2 (1-cpi)^n_2$

then we get

$binomn_1y_1 binomn_2y_2 (1-pi)^n_1 exp Big((y_1+y_2) log dfracpi1-piBig) exp Big(y_2log dfrac(1-pi)c1-cpiBig) (1-cpi)^n_2$

Where the canonical statistic is $u = Y_1 +Y_2$ and $v = Y_2$. I don't see how I can get $v = Y_1$ and $u = Y_1 + Y_2$ . Can anyone give me a hint?

probability-theory binomial-distribution

share|cite|improve this question

edited Mar 22 at 14:24

Cettt

2,010623

asked Mar 22 at 8:18

J.doeJ.doe

515

$endgroup$

add a comment | 

0

$begingroup$

A statistical model for a data set y is an exponential family , with canonical parameter vector $theta= (theta_1,theta_2,..theta_k)$ and canonical statistic $t(y) = (t_1(y),t_2(y),..t_k(y))$ if $f$ has structure

$$f(y;theta) = a(theta)h(y) exp(theta^T t(y)).$$

Now,consider two binomially distributed random variables $Y_1 sim Bin(n_1,pi)$ and $Y_2 sim Bin(n_2,cpi)$ prove that the canonical statistic $t(Y_1,Y_2) = (v,y)^T$ is $v = Y_1,u = Y_1 + Y_2$

$binomn_1y_1 binomn_2y_2 pi^y_1(1-pi)^n_1-y_1 (cpi)^y_2(1-cpi)^n_2-y_2 $

The only thing I can come up with is the form:

$binomn_1y_1 binomn_2y_2 (1-pi)^n_1 Big( dfracpi1-piBig)^y_1+y_2 Big(dfrac(1-pi)c1-cpiBig)^y_2 (1-cpi)^n_2$

then we get

$binomn_1y_1 binomn_2y_2 (1-pi)^n_1 exp Big((y_1+y_2) log dfracpi1-piBig) exp Big(y_2log dfrac(1-pi)c1-cpiBig) (1-cpi)^n_2$

Where the canonical statistic is $u = Y_1 +Y_2$ and $v = Y_2$. I don't see how I can get $v = Y_1$ and $u = Y_1 + Y_2$ . Can anyone give me a hint?

probability-theory binomial-distribution

share|cite|improve this question

edited Mar 22 at 14:24

Cettt

2,010623

asked Mar 22 at 8:18

J.doeJ.doe

515

$endgroup$

add a comment | 

$begingroup$

A statistical model for a data set y is an exponential family , with canonical parameter vector $theta= (theta_1,theta_2,..theta_k)$ and canonical statistic $t(y) = (t_1(y),t_2(y),..t_k(y))$ if $f$ has structure

$$f(y;theta) = a(theta)h(y) exp(theta^T t(y)).$$

Now,consider two binomially distributed random variables $Y_1 sim Bin(n_1,pi)$ and $Y_2 sim Bin(n_2,cpi)$ prove that the canonical statistic $t(Y_1,Y_2) = (v,y)^T$ is $v = Y_1,u = Y_1 + Y_2$

$binomn_1y_1 binomn_2y_2 pi^y_1(1-pi)^n_1-y_1 (cpi)^y_2(1-cpi)^n_2-y_2 $

The only thing I can come up with is the form:

$binomn_1y_1 binomn_2y_2 (1-pi)^n_1 Big( dfracpi1-piBig)^y_1+y_2 Big(dfrac(1-pi)c1-cpiBig)^y_2 (1-cpi)^n_2$

then we get

$binomn_1y_1 binomn_2y_2 (1-pi)^n_1 exp Big((y_1+y_2) log dfracpi1-piBig) exp Big(y_2log dfrac(1-pi)c1-cpiBig) (1-cpi)^n_2$

Where the canonical statistic is $u = Y_1 +Y_2$ and $v = Y_2$. I don't see how I can get $v = Y_1$ and $u = Y_1 + Y_2$ . Can anyone give me a hint?

probability-theory binomial-distribution

share|cite|improve this question

edited Mar 22 at 14:24

Cettt

2,010623

asked Mar 22 at 8:18

J.doeJ.doe

515

$endgroup$

share|cite|improve this question

edited Mar 22 at 14:24

Cettt

2,010623

asked Mar 22 at 8:18

J.doeJ.doe

515

share|cite|improve this question

edited Mar 22 at 14:24

Cettt

2,010623

asked Mar 22 at 8:18

J.doeJ.doe

515

edited Mar 22 at 14:24

Cettt

2,010623

edited Mar 22 at 14:24

Cettt

2,010623

asked Mar 22 at 8:18

J.doeJ.doe

515

asked Mar 22 at 8:18

J.doeJ.doe

515