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Given an i.i.d. sample $X = (x_{1}, \dots, x_{n}) \sim N(\mu, 1)$. I have been asked to show that the likelihood of $\mu$ based on the whole sample is proportional to the likelihood based on $\bar{x}$ (the sample mean) alone. I'm a little confused by the phrasing of this question. I was hoping the perhaps somebody could elaborate on what this question is asking.

Also, as an aside, given a random sample $X = (x_{1}, \dots, x_{n}) \sim N(\mu, \sigma^{2})$, how can one estimate the $\text{Pr(X > x)}$?

Can the MLEs for $\mu$ and $\sigma^{2}$ be used in normal CDF to estimate $\text{Pr(X > x)}$?

where

$\hat{\mu} = \bar{x} = \frac{\sum_{i}x_{i}}{n}$

and

$\hat{\sigma}^{2} = \frac{\sum_{i}(x_{i}-\hat{\mu})^{2}}{n}$

hence

$\text{Pr}(X > x) = 1 - F(x)$

where $F(x)$ is the CDF evaluated at $x$.

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2  
Write down the log likelihood: obviously it is a sum of squares of things. Algebraically attempt to rewrite that in terms of $\bar{x}$; if you're successful, you will see that the likelihood depends on the sample only through the sample mean and you will see precisely how it depends on that mean. A little reflection should then indicate how to answer your "aside" question. Your third question, when inverted (given $\alpha$, to find an $x$ for which $\Pr(X\gt x)=\alpha$) is a tolerance limit; there are several kinds that cope variously with uncertainties in $\mu$ and $\sigma$. –  whuber Feb 26 '13 at 19:32
    
@whuber his third question is just about the estimate of $\Pr(X>x)$ for a given $x$, I think –  Stéphane Laurent Feb 26 '13 at 21:58
    
Your comment, @Stéphane, makes me realize there is an ambiguity there. I had taken the last question to be about the mean of $X$, but maybe it's not. If not, it's natural to interpret "$X\gt x$" as being a conjunction of events $X_1\gt x_1,X_2\gt x_2,\ldots,X_n\gt x_n$ (which is what such notation means in mathematical communities that deal frequently with complicated sets of inequalities). –  whuber Feb 26 '13 at 22:15
    
@StéphaneLaurent is correct - the third question is about the estimate of $\text{Pr(X>x)}$ for a given $x$. My apologies for the sloppy notation and any ambiguity caused. –  Will Clyne Feb 26 '13 at 22:25
1  
You have written $L(\mu \mid x_1, \ldots, x_n) = \exp\left(-\frac{n}{2}(\mu-\bar{x})^{2}\right) \times \text{"something"}$ with "something" not depending on $\mu$ so that's the answer to your initial question. –  Stéphane Laurent Feb 28 '13 at 11:16
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