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analytical distribution of the maximum likelihood estimator for a uniform distribution

0

$begingroup$

Obviously the MLE of $theta$ for a distribution $X_1, X_2, dots, X_n sim Uniform(0,theta)$ is $hattheta = max(X_1, X_2,dots,X_n)$

Now, assume $theta = 1$. If you take repeated samples with $n=50$. What would the distribution of $hattheta$ be?

I assume it would be:
$$f(x) = P(hattheta = x) = P(X_1 le x) P(X_2 le x) dots P(X_50 le x) = x^50$$

given that $x≤1$ always since

However, if you integrate this distribution, it does not sum to 1:
$$int_0^1x^50dx = frac151$$

so, would the "true" pdf be $$f(x) = 51x^50$$
or would it be a different function altogether?

probability-distributions self-learning uniform-distribution maximum-likelihood

share|cite|improve this question

edited Mar 31 at 13:59

lstbl

asked Mar 31 at 13:48

lstbllstbl

19913

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You need to review the definition of a probability density function. And $P(X_1le x)=fracxtheta$ for all $0<x<theta$.
  $endgroup$
  – StubbornAtom
  Mar 31 at 13:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I stated that $theta = 1$, so $P(X_1 le x) = fracxtheta = x$.
  $endgroup$
  – lstbl
  Mar 31 at 13:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Okay. But the blunder is '$f(x)=P(hattheta=x)=cdots$' assuming $f$ is the pdf of $hattheta$. It is actually the cdf $F(x)=P(hatthetale x)=(P(X_1le x))^n$. Now find pdf from cdf.
  $endgroup$
  – StubbornAtom
  Mar 31 at 14:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  AHHHHHH! Totally makes sense. I guess I was thinking about this a little backwards. What is the probability that $hattheta = x$... well I guess that probability is 0 for any continuous distribution function. However you CAN make a statement about $P(X≤x)$ for a continuous function.
  $endgroup$
  – lstbl
  Mar 31 at 14:08
  

  
  
  
  

add a comment | 

0

$begingroup$

Obviously the MLE of $theta$ for a distribution $X_1, X_2, dots, X_n sim Uniform(0,theta)$ is $hattheta = max(X_1, X_2,dots,X_n)$

Now, assume $theta = 1$. If you take repeated samples with $n=50$. What would the distribution of $hattheta$ be?

I assume it would be:
$$f(x) = P(hattheta = x) = P(X_1 le x) P(X_2 le x) dots P(X_50 le x) = x^50$$

given that $x≤1$ always since

However, if you integrate this distribution, it does not sum to 1:
$$int_0^1x^50dx = frac151$$

so, would the "true" pdf be $$f(x) = 51x^50$$
or would it be a different function altogether?

probability-distributions self-learning uniform-distribution maximum-likelihood

share|cite|improve this question

edited Mar 31 at 13:59

lstbl

asked Mar 31 at 13:48

lstbllstbl

19913

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You need to review the definition of a probability density function. And $P(X_1le x)=fracxtheta$ for all $0<x<theta$.
  $endgroup$
  – StubbornAtom
  Mar 31 at 13:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I stated that $theta = 1$, so $P(X_1 le x) = fracxtheta = x$.
  $endgroup$
  – lstbl
  Mar 31 at 13:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Okay. But the blunder is '$f(x)=P(hattheta=x)=cdots$' assuming $f$ is the pdf of $hattheta$. It is actually the cdf $F(x)=P(hatthetale x)=(P(X_1le x))^n$. Now find pdf from cdf.
  $endgroup$
  – StubbornAtom
  Mar 31 at 14:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  AHHHHHH! Totally makes sense. I guess I was thinking about this a little backwards. What is the probability that $hattheta = x$... well I guess that probability is 0 for any continuous distribution function. However you CAN make a statement about $P(X≤x)$ for a continuous function.
  $endgroup$
  – lstbl
  Mar 31 at 14:08
  

  
  
  
  

add a comment | 

$begingroup$

Obviously the MLE of $theta$ for a distribution $X_1, X_2, dots, X_n sim Uniform(0,theta)$ is $hattheta = max(X_1, X_2,dots,X_n)$

Now, assume $theta = 1$. If you take repeated samples with $n=50$. What would the distribution of $hattheta$ be?

I assume it would be:
$$f(x) = P(hattheta = x) = P(X_1 le x) P(X_2 le x) dots P(X_50 le x) = x^50$$

given that $x≤1$ always since

However, if you integrate this distribution, it does not sum to 1:
$$int_0^1x^50dx = frac151$$

so, would the "true" pdf be $$f(x) = 51x^50$$
or would it be a different function altogether?

probability-distributions self-learning uniform-distribution maximum-likelihood

share|cite|improve this question

edited Mar 31 at 13:59

lstbl

asked Mar 31 at 13:48

lstbllstbl

19913

$endgroup$

share|cite|improve this question

edited Mar 31 at 13:59

lstbl

asked Mar 31 at 13:48

lstbllstbl

19913

share|cite|improve this question

edited Mar 31 at 13:59

lstbl

asked Mar 31 at 13:48

lstbllstbl

19913

asked Mar 31 at 13:48

lstbllstbl

19913

asked Mar 31 at 13:48

lstbllstbl

19913