Recently I was asked the following problem:
Inside the equilateral triangle ABC with area 1 there was randomly chosen some point M. Calculate the expected value of an area of the triangle ABM.
Here is my solution:
Let $X$ be a random variable that indicates the distance between point $M$ and side $AB$(it is easy to show that $X$ is also a height of a triangle $ABM$ from point $M$). Obviously, X is uniformly distributed between $0$ and $h$, where $h$ is a distance between side $AB$ and $C$ (height of $ABM$ when $M \equiv C$). Let $a$ be the length of a side of $ABC$.
$PDF$ of X is given by $P(X) = \frac{1}{h-0} 1_{(0, h)}=\frac{1}{h} 1_{(0, h)}$, because X is uniformly distributed
Then the expected value can be written as: $$ E[X] = \int_\mathbb{R}xP(x)dx \\ =\int_0^1\frac{1}{h}xdx=\frac{1}{h} \big[ \frac{x^2}{2}\big] \big|_0^h =\frac{h}{2} $$
Suppose $Y$ is a random variable indicates the area of the triangle ABC. Then $Y=\frac{1}{2}aX$. Hence, we got:
$$ E[Y]=E[\frac{1}{2}aX]=\frac{1}{2}aE[X]=\frac{1}{2} \frac{ha}{2} = \frac{1}{2} Area_{ABC}=\frac{1}{2} $$
But the keys says that $E[Y]=\frac{1}{3}$ and I believe it because expected values of an areas of the triangles $ABM$, $BCM$, $ACM$ intuitively should be the same. Therefore $$E[Area_{ABM}]=E[Area_{BCM}]=E[Area_{ACM}]=\frac{1}{3}Area_{ABC}=\frac{1}{3}$$
Then I googled it and found the exact my problem on math.stackexchange. But I didn't find there why I got different answers. So, please do not say that my question is duplicate.
Now, I want to understand why does the way with evaluating expected value by it's definition give the wrong answer? I'm stuck with it and need a help.
Your claim that
is a false claim.
$X$ is more likely to be close to $0$ than close to $h$.
For example, $X$ is $3$ times more likely to be in the interval $(0,{\large{\frac{h}{2}}})$ than in the interval $({\large{\frac{h}{2}}},h)$.