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What does $f(x\mid\theta)$ denote?

Isn't it interprets that- "here $\theta$ is known then what is the distribution of X for the known $\theta$?"

Then why in the book which i am reading wrote that -- "the function $f(.\mid\theta)$ is assumed known except for $\theta$?"

If $\theta$ is known then why will the function be unknown for $\theta$?

Again, if the statement of the book is true [i know, it is true but I am not understanding] then how is $f(.|\theta)$ known at every point except for $\theta$?

Again, why the is book using the notation $f(.|\theta)$ instead of $f(x|\theta)$ ? Why is the dot(.) being used instead of $x$? that is what does $f(.|\theta)$ symbolize?

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A dot represents what's called a placeholder variable. It's a space where I can put a number. $x$ represents an actual number.

So the ordinary density $f(x)$ is the value of the density at $x$. We might not know $x$ but technically it still has a value. Which means $f(x)$ is a number.

But the density itself is a function that is take a number (perhaps $x$) and follow a set of instructions to get another number.

$f(\cdot)$ represents the set of instructions itself, not the result of those instructions when carried out at $x$. That's why we've left a little dot, to tell you where to put an $x$.

For example I can define a function on two variables by setting $$g(x,y) = \frac xy$$

That's a statement that's true for every pair $x,y$ ($y\neq 0$). It's not the function itself, it's just something I'm writing to help you know what I mean by the function $g = g(\cdot,\cdot)$.

Of course that function already had a name it's usually denoted by $$g=\, \frac{\,\cdot\,}\cdot$$

So for the conditional density $f(\cdot|\theta)$ is a function of one variable, if you give it a value $x$ it gives you a value $f(x|\theta)$.

The reason the book uses a dot here is because if it just wrote $f$ it could be referring to the function $f(\cdot|\cdot)$ on two variables, or even the function $f(x,\cdot)$ which takes $\theta\mapsto f(x,\theta)$.

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  • $\begingroup$ ~Thank you very much. your explanation is really very helpful. $\endgroup$ – ABC May 5 '13 at 17:57

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