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$n$ random variables or a random sample of size $n$ $\quad X_1,X_2,\ldots,X_n$ assume a particular value $\quad x_1,x_2,\ldots,x_n$ .

  • What does it mean? The set $\quad x_1,x_2,\ldots,x_n$ constitutes only a single value? or, $\quad x_1,x_2,\ldots,x_n$ are n values , that is, $X_1$ assumes the value $x_1$, $X_2$ assumes the value $x_2$,and so on?

  • Why Likelihood Function is a function of parameter $\theta$? Why not is the function of random variables $\quad X_1,X_2,\ldots,X_n$?

  • Let $n=1$ and $X_1$ has the normal density with mean, $\mu=6$ and variance $\sigma^2=1.$ Then the value of $X_1$ which is most likely to occur is $X_1=6.$ How it has been computed? I have thought in the way that since $X_1$ is a random variable it can assume the values $x_1,\ldots,x_n$ [Then how $\quad X_1,X_2,\ldots,X_n$ can assume $\quad x_1,x_2,\ldots,x_n$?] and since the mean of the random variable is $6$ so The value which is most likely to occur is $6$. Is it the real process ?

Michael Hardy
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time
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1 Answers1

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In your first question, what is meant is that $X_1$ assumes the value $x_1$ and $X_2$ assumes the value $x_2$, etc. That may be expressed by saying that the tuple or vector $(X_1,\ldots,X_n)$ assumes the value $(x_1,\ldots,x_n)$.

To your third question, one must recognize first that the probability that $X_1$ assumes any particular value is $0$, since this is a continuous distribution. But the probability that $X_1$ is in any interval centered at $6$ is greater than the probability that $X_1$ is in any equally long interval centered anywhere else. Just look at the bell-shaped curve and remember that the probability is the area under the curve.

The reason that the likelihood is a function of $\theta$ with $x_1,\ldots,x_n$ fixed is just that the term "likelihood function" is conventionally used to refer to that function. In the case of the family of all normal distributions, this would be a function of the pair $(\mu,\sigma)$ or of the pair $(\mu,\sigma^2)$.

Michael Hardy
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  • According to my first question your answer is correct. But i think my first question is incorrect because if $X_1$ is a random variable then it can't assume a particular value rather it will assume a number of values such as $x_1,...,x_n$. I think the question will be correct if the statement is: "The Likelihood Function gives the likelihood that $n$ random variables or a random sample of size $n$ $\quad X_1,X_2,\ldots,X_n$ assume a particular value $\quad x_1,x_2,\ldots,x_n$ ." Is it so? – time Jun 13 '13 at 03:16
  • @MichealHardy Would you explain "the family of all normal density"? Is it $X_1\sim N(\mu,\sigma^2)$,$X_2\sim N(\mu,\sigma^2),\ldots, X_n\sim N(\mu,\sigma^2)$ ? That is, $X_1,\ldots, X_n$ independent and identically distributed as Normal with mean,$\mu$ and variance,$\sigma^2$? – time Jun 13 '13 at 03:42
  • @tree : Yes. ${}\qquad{}$ – Michael Hardy Jun 14 '13 at 00:55
  • Thank you very much. I have posted another question about why likelihood function is "joint density" of $n$ random variables. Not why likelihood function is a function of $\theta$. But most readers did not understand the difference of the question. http://math.stackexchange.com/questions/419287/likelihood-function-for-the-uniform-density?noredirect=1#comment896256_419287 – time Jun 14 '13 at 02:28