I have a continuous of $X$, $f_X$. Now I want to find the argument that maximizes $f_X$, i.e., $\arg \max_x f_X$. I know that for a continuous distribution, the probability for $X = x$ is 0. Thus, I think we can replace $\arg \max_x f_X$ with simply $\max_x f_X$. This $x$ will maximize $f_X$. Is this a correct conclusion?
Asked
Active
Viewed 36 times
0
-
$f_X$ (which I assume is a probability density) is just a function, and $\max_x f_X$ and $\arg\max_x f_X$ have their usual meaning: the argmax is the value of $x$ that makes $f_X(x)$ largest, while the $\max$ is the aforementioned largest value of $f_X(x)$. [So far, this has nothing to do with probability.] The issue with probability densities is that you cannot interpret the argmax of the density as the "most probable outcome." – angryavian Sep 30 '20 at 05:57
-
I understand. So, let's say I get the maximum value of $f_X$ for $x = x^$. Then use this value to get $\max_x f_X$. To get the $x = x^$, I can compute the first derivate and set it to 0. This is the right approach, right? – Bikas Sep 30 '20 at 06:00