Timeline for Maximum likelihood estimate for uniform distribution
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 5, 2015 at 19:07 | comment | added | Silverfish | @user149705 "I don't understand why we have to consider densities" - because for continuous distributions, the probability of being any value in particular (e.g. $P(X=12)$) is zero. So we don't learn a lot by looking at probabilities. If $X\sim U(-1,1)$ we'd like to say $X=2$ is "impossible", and $X=0.2$ is "just as likely" as $X=-0.7$. But $P(X=2)=0$, $P(X=0.2)=0$ and $P(X=-0.7)=0$. More usefully, $f(2)=0$, $f(0.2)=0.5$ and $f(-0.7)=0.5$. | |
| Jan 5, 2015 at 16:28 | answer | added | Silverfish | timeline score: 5 | |
| Jan 5, 2015 at 16:28 | comment | added | user149705 | AlecosPapadopoulos : Why it is equivalent to consider maximizing the density against maximizing the probability to observe the samples. Silverfish : The probability to observe those sample given the parameter is 0. I don't understand why we have to consider densities here ( even if in the definition it says to ). DilipSarwate : Yes I can figure out the result of maximizing the likelihood function, this is not totally my question but why do we actually consider densities here. | |
| Jan 5, 2015 at 14:18 | comment | added | Dilip Sarwate | Hint: the likelihood of the observation $(X_1=x_1,X_2=x_2; a)$ has value $\frac{1}{(2a)^2}$ for some choices of $a$ and value $0$ for other choices of $a$. Can you figure out what these choices are (Subhint: the answer will depend on $x_1$ and $x_2$ in some way)? Can you sketch the likelihood as a function of $a$? Where do you think it might attain maximum value? (Subhint: no calculus need be harmed in answering this last query; just looking at the sketch should suffice.) | |
| Jan 5, 2015 at 11:29 | comment | added | Silverfish | "The likelihood of the samples say 12 and 30 is defined by the probability to observe those sample given the parameters" I suggest you go back and check your definition. For continuous functions does it really say "probability"? | |
| Jan 5, 2015 at 10:22 | comment | added | Alecos Papadopoulos | Which is mathematically equivalent to which? | |
| Jan 5, 2015 at 7:14 | review | First posts | |||
| Jan 5, 2015 at 7:37 | |||||
| Jan 5, 2015 at 7:14 | history | asked | user149705 | CC BY-SA 3.0 |