Kindly see the boldened sentence below. What probability concept does "the most likely value" refer to? How do I symbolize it?

I don't think the author's referring to Expected Value, because it's additive.

      The principle of additivity is so intuitively appealing that it’s easy to think it’s obvious. But, just like the pricing of life annuities, it’s not obvious! To see that, substitute other notions in place of expected value and watch everything go haywire. Consider:
      The most likely value of the sum of a bunch of things is the sum of the most likely values of each of the things. [emphasis mine]
      That’s totally wrong. Suppose I choose randomly which of my three children to give the family fortune to. The most likely value of each child’s share is zero, because there’s a two in three chance I’m disinheriting them. But the most likely value of the sum of those three allotments—in fact, its only possible value—is the amount of my whole estate.

Ellenberg, How Not to Be Wrong (2014), page 213.

  • $\begingroup$ the point that maximizes the density, i guess? $\endgroup$ Jun 2, 2021 at 23:45
  • $\begingroup$ Where is the confusion? If you have a list of possible values with various (unequal) probabilities, the most likely value is the one with the greatest probability. $\endgroup$
    – lulu
    Jun 2, 2021 at 23:46
  • 1
    $\begingroup$ For another numerical example, suppose you have a loaded die which comes up $1$ with probability $\frac 16+\epsilon$, and takes all the other values with probability $\frac 16-\frac {\epsilon}5$, for very small $\epsilon$. Then of course the most likely value is $1$. But the most likely value of the sum of two rolls is still $7$, not $2$. $\endgroup$
    – lulu
    Jun 2, 2021 at 23:49
  • $\begingroup$ The writer's heuristic sort of holds true for independent random variables. $\endgroup$
    – K.defaoite
    Jun 3, 2021 at 0:51

2 Answers 2


It sounds like you're trying to describe a situation in which $\operatorname{mode}(X_1 + \cdots +X_n) = \operatorname{mode} X_1 + \cdots + \operatorname{mode} X_n$.

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    $\begingroup$ But, to be clear, mode can have multiple values... $\endgroup$ Jun 3, 2021 at 1:04
  • 3
    $\begingroup$ Absolutely, just trying to put in symbols what I think the original post was saying. Whether it makes sense or not is a different thing. +1. $\endgroup$ Jun 3, 2021 at 3:28

You can only really define the “most likely” value for some random variable. If $f$ is a pdf for a random $X,$ then the most like value of $X$ is the “mode” of $X,$ a value $x_0$ such that:

$$f(x_0)=\sup_x f(x).$$

But the mode is not necessarily unique. And even if $X_1$ and $X_2$ have a unique mode, $X_1+X_2$ might not.

I suppose, if $f$ is continuous, you could define $M_f=\sup_x f(x),$ then define, for $M<M_f,$ $E_{f,M} =\{x\mid f(x)>M\}$ and define our “most like value” as:

$$\begin{align}\mathcal A(X)&=\lim_{M\to {M_f}^-}\frac{\int_{E_{f,M}}xf(x)\,dx}{\int_{E_{f,M}}f(x)\,dx} \\&=\lim_{M\to {M_{f}}^{-}}E\left(X\mid X\in E_{f,M}\right)\end{align}$$

That will be, roughly, a weighted average when there are multiple values for the mode, where the weights prefer the ones where $f$ converges more slowly to $\sup f.$ In particular, if there is one mode, $A(X)$ will be that mode.

This definition satisfies some useful rules.

$\mathcal A(aX+b)=a\mathcal A(X)+b$ when $a,b$ are (constant) real numbers.

If $X$ is a an even distribution ($f(x)=f(-x)$) then $\mathcal A(X)=E(X)=0.$

This definition works for some $f$ unbounded, for example $f(x)=\frac{1}{2\sqrt{x}}$ in $(0,1].$ Then $M\to +\infty.$

If $X_1,\dots,X_n$ are real continuous random variables, then we would want:

$$\mathcal A(X_1+X_2+\cdots+X_n)=\mathcal A(X_1)+\mathcal A(X_2)+\cdots +\mathcal A(X_n).$$

This is true if the $X_i$ are normal, or more generally if $X_i-E(X_i)$ is even.

But this won’t be true in general.

  • $\begingroup$ Thanks! My 10 year old and I will need to re-read your post, after we learn calculus LOL. $\endgroup$
    – user53259
    Jun 3, 2021 at 5:43

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