j*********1
发帖数: 15 | 1 Need help on this problem
Test
null: theta=theta.a
alternative theta=theta.b (b
theta is the parameter of exponential distribution
have X1,... Xk+1 (each follow the exponential distribution) at different
time interval t1,...tk (equal) and tk+1=infinite. so X1,... Xk+1 follow a
multinomial distribution with pi determined by pdf theta from ti-1 to ti.
Now consider use likelihood ratio test (used MLE before and want to compare)
Now we have lnLR=sum of [xi*ln(p(theta.a/theta.b)]
how to approximate this distribution for large sample size case?
I computed the expectation and variance of the right hand side and thought
it may be a asymptotic normal but guess it is not right. Any suggestion? | m*****O
发帖数: 3558 | 2 didnt get ur question. but logLR is always asymp chi squared ba
【在 j*********1 的大作中提到】
: Need help on this problem
: Test
: null: theta=theta.a
: alternative theta=theta.b (b: theta is the parameter of exponential distribution
: have X1,... Xk+1 (each follow the exponential distribution) at different
: time interval t1,...tk (equal) and tk+1=infinite. so X1,... Xk+1 follow a
: multinomial distribution with pi determined by pdf theta from ti-1 to ti.
: Now consider use likelihood ratio test (used MLE before and want to compare)
: Now we have lnLR=sum of [xi*ln(p(theta.a/theta.b)]
| j*********1
发帖数: 15 | 3 对于chi-square,alternative hypothesis是得pi是MLE吧（Xi/N),但是这里是已经定了
pi（theta.b和time interval决定)。Hogg的书上有asymp MLE到normal的，不知道是不
是也适用于这里的LR
【在 m*****O 的大作中提到】
: didnt get ur question. but logLR is always asymp chi squared ba
| n*****n
发帖数: 3123 | 4 have X1,... Xk+1 (each follow the exponential distribution) at different
time interval t1,...tk (equal) and tk+1=infinite.
你这个是什么意思，没看明白 | j*********1
发帖数: 15 | 5 X1 is the # of items at time interval t0 to t1
Xi follow exponential dist
【在 n*****n 的大作中提到】
: have X1,... Xk+1 (each follow the exponential distribution) at different
: time interval t1,...tk (equal) and tk+1=infinite.
: 你这个是什么意思，没看明白
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