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# log likelihood; statistics Watch

1. not sure why the the 2nd line eq. says 'proportional to' the right hand bit, what happens to the {2Ø(1-Ø)}^n2 on the left hand side? because surely if the LHS wasn't proportioned, the log-likelihood would be different if the {2Ø(1-Ø)}^n2 bit was included?
2. (Original post by rextra)
not sure why the the 2nd line eq. says 'proportional to' the right hand bit, what happens to the {2Ø(1-Ø)}^n2 on the left hand side? because surely if the LHS wasn't proportioned, the log-likelihood would be different if the {2Ø(1-Ø)}^n2 bit was included?
For the bit in blue, everything has been included apart from the and hence the proportionality.
3. (Original post by ghostwalker)
For the bit in blue, everything has been included apart from the and hence the proportionality.
yes however can that be something that can just be done, ie can just be avoided? isn't that important when taking logs?
4. (Original post by rextra)
yes however can that be something that can just be done, ie can just be avoided? isn't that important when taking logs?
When you take logs, it would appear as +log(constant) and when you differentiated it would vanish.
5. (Original post by ghostwalker)
When you take logs, it would appear as +log(constant) and when you differentiated it would vanish.
oh I misread that, cheers x
6. (Original post by ghostwalker)
When you take logs, it would appear as +log(constant) and when you differentiated it would vanish.
forgive me for the endless questions, but please can ya sort me out with this badboy?

I know how to integrate it normally but am keen on understanding my lecturer's method since it seems shorter, if one gets it anyway. He said he used the substitution of y=λ(x-a) . I don't get how on the RHS there's a 1/λ outside the integral and how has the limit changed from a-->0? He mentioned solving it by comparing it with the exponential dist. where λ=1 or something?
7. (Original post by rextra)
forgive me for the endless questions, but please can ya sort me out with this badboy?

I know how to integrate it normally but am keen on understanding my lecturer's method since it seems shorter, if one gets it anyway. He said he used the substitution of y=λ(x-a) . I don't get how on the RHS there's a 1/λ outside the integral and how has the limit changed from a-->0? He mentioned solving it by comparing it with the exponential dist. where λ=1 or something?
have you covered integration by substitution - this is just a standard example.

Putting means we can replace with

We need to change the limits, so x=a corresponds to y=0. And , so you replace dx with
8. (Original post by davros)
have you covered integration by substitution - this is just a standard example.

Putting means we can replace with

We need to change the limits, so x=a corresponds to y=0. And , so you replace dx with
yeah just had to re-visit integration, cheers. But what I really want to know is what he did after that. In his lectures he mentioned solving the integral obtained in a short cut by using the exponential distribution with =1 and then treating it like a probability density function or something?
9. (Original post by rextra)
yeah just had to re-visit integration, cheers. But what I really want to know is what he did after that. In his lectures he mentioned solving the integral obtained in a short cut by using the exponential distribution with =1 and then treating it like a probability density function or something?
I'm not sure - it may be possible to relate it to another distribution, but to be honest you can evalute that integral directly - is easy to integrate, and is just a simple integration by parts
10. (Original post by ghostwalker)
For the bit in blue, everything has been included apart from the and hence the proportionality.
just went through this again and I can't figure out what you've done with θ^2n1 *[2θ(1-θ)]^n2? Please elaborate? thanks in advance mate
11. (Original post by rextra)
just went through this again and I can't figure out what you've done with θ^2n1 *[2θ(1-θ)]^n2? Please elaborate? thanks in advance mate

Expands to

And when grouped with the other terms, using the rules of indices, yields the desired result.
12. (Original post by ghostwalker)

Expands to

And when grouped with the other terms, using the rules of indices, yields the desired result.
you're the best, thanks!

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