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I've been trying to get a better understanding of distributions. So far I understand how we get the formulas for mean and variance (by looking at the derivative of the moment generating function). However, I'm confused as to the general process of finding the mode.

I understand the process laid out here, but I'm wondering why one would even look at the $k+1$ and $k$ terms in the first place.

emka
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The question you refer to in the link seems to have been the subject of some bickering. Anyway, I gather that you're asking how to find the mode of a distribution in general, not just for the binomial distribution.

The mode is the outcome(s) which arises most frequently. This is easy to understand in a discrete random variable. If $X$ is a random variable which takes values in $\Omega$ then the $mode$ is the value $x \in \Omega$ for which $Pr[X= x]$ is maximised, in other words the point $x$ at which the p.m.f. $p(x)$ is a maximum. There may be many such points, in which case there is more than one $mode$. These points are not necessarily always next to each other.

With a continuous random variable, the $mode$ is the point(s) $x$ at which the density function $f(x)$ is a maximum. This can usually be found by differentiating the density function to find the points where the derivative is zero and then, importantly, also checking whether such points are actually maxima.

Stanley
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    Adding onto Walter's answer: The exceptions to his "usually" are (for example) instances in which the function is not differentiable at the peak. For instance, the PDF $f_X(x) = 1-|x|$ for $-1 \leq x \leq 1$, $f(x) = 0$ otherwise has a peak (and $X$ has a mode) at $x = 0$, but $f_X(x)$ is not differentiable there. That kind of situation is where you must examine the PDF directly, and not rely on setting the derivative equal to zero. – Brian Tung May 27 '15 at 19:52