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log likelihood; statistics

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?logsssss.jpg
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?logsssss.jpg


For the bit in blue, everything has been included apart from the 2n2\displaystyle 2^{n_2} and hence the proportionality.
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Original post by ghostwalker
For the bit in blue, everything has been included apart from the 2n2\displaystyle 2^{n_2} and hence the proportionality.


yes however can that be something that can just be done, ie can 2n2\displaystyle 2^{n_2} just be avoided? isn't that important when taking logs?
Original post by rextra
yes however can that be something that can just be done, ie can 2n2\displaystyle 2^{n_2} 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.
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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
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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? Untitled.jpg

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?
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Original post by rextra
forgive me for the endless questions, but please can ya sort me out with this badboy? Untitled.jpg

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 y=λ(x−a)y = \lambda(x-a) means we can replace λx\lambda x with y+λay + \lambda a

We need to change the limits, so x=a corresponds to y=0. And dy/dx=λdy/dx = \lambda, so you replace dx with λ−1dy\lambda^{-1}dy
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Original post by davros
have you covered integration by substitution - this is just a standard example.

Putting y=λ(x−a)y = \lambda(x-a) means we can replace λx\lambda x with y+λay + \lambda a

We need to change the limits, so x=a corresponds to y=0. And dy/dx=λdy/dx = \lambda, so you replace dx with λ−1dy\lambda^{-1}dy


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 λ\lambda=1 and then treating it like a probability density function or something?
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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 λ\lambda=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 - e−ye^{-y} is easy to integrate, and ye−yye^{-y} is just a simple integration by parts :smile:
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Original post by ghostwalker
For the bit in blue, everything has been included apart from the 2n2\displaystyle 2^{n_2} 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
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


[2θ(1−θ)]n2[2\theta (1-\theta)]^{n_2}

Expands to 2n2θn2(1−θ)n22^{n_2}\theta^{n_2} (1-\theta)^{n_2}

And when grouped with the other terms, using the rules of indices, yields the desired result.
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Original post by ghostwalker
[2θ(1−θ)]n2[2\theta (1-\theta)]^{n_2}

Expands to 2n2θn2(1−θ)n22^{n_2}\theta^{n_2} (1-\theta)^{n_2}

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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