Most probable value vs maximum of the distribution [closed]

Question

Given a distribution p(x), there are two things that can be calculated.

1. Value of x for which p(x) is maximum.
2. Most probable value of x weighted over p(x).

Would these two values of x be the same?

Answer 1

No.

To make it easier for comprehension, we may consider the discrete case, where 'p(x) is maximum' refers to the event that takes place most, while 'Most probable value of x weighted over p(x).' (expectation) refers to the average value. In a extreme case if you have a dice with four '0's and two '1's, the value of x for which p(x) is maximum is clearly $0$ and the most probable value of x weighted over p(x) is $1/3$, which is even not a value you would ever see in any cases.

Also, it's actually a bit confusing when you say "Most probable value of x weighted over p(x)", for the term "Most probable value" itself means value of x for which p(x) is maximum. I guess you are trying to refer to the expectation.

Answer 2

My answer might not be a full-fledge solution but I think it should still help.

Clearly, the most probable value of x weighted over $P(x)$ is an expected value $E[X]$. Now, assume that what you claim is true. Let's say that there is a discrete probability where $P(x=12)=0.2$ and $P(x=2)=0.8$ Clearly, the value of $x$ for which $P(x)$ is maximum is $2$ but the expected value is $E[X]=\sum^2_{i=1}P(X=x_i)x_i=P(x=2)*2+P(x=12)*12=0.8*2+0.2*12=4.0$. Well, $12\ne4$ and I would argue that even though $4.0$ is closer to $2$ than $12$, you can't just say that these values are the same

Answer 3

The expected value and most probable value can but need not coincide. Consider a random variable $X$ defined by $P(X=0)=0.9$ and $P(X=1)=0.1$. The most probable value is $0$, yet the expected value is $0.1$.

If your idea of “most probable value weighted over the distribution” is the expected value, then, no, these values need not be equal.