# Expected value vs. most probable value (mode)

The expected value of a distribution $$f(x)f(x)$$ is the mean, that is the weighted average value
$$E[x]=\int_{-\infty}^{+\infty} x \, \, f(x) dxE[x]=\int_{-\infty}^{+\infty} x \, \, f(x) dx$$

The most likely value is the mode, that is the most probable value.

However do we expect somehow to see $$E[x]E[x]$$ a lot of times? Quoting from here:

If the outcomes $$x_ix_i$$ are not equally probable, then the simple average must be replaced with the weighted average, which takes into account the fact that some outcomes are more likely than the others. The intuition however remains the same: the expected value of $$xx$$ is what one expects to happen on average.

I cannot understand what does “happen on average” means, does this mean that, for istance, taking a measure a lot of time I expect to see $$E[x]E[x]$$ more than other values of $$xx$$? But isn’t this the definition of mode?

So how to interpret the statement? And what is the probabilistic meaning of $$E[x]E[x]$$?

I would also like to show an example where I do get confused. Studying $$\chi^2\chi^2$$ distribution I learned that the mode
is $$\chi^2_{mode}=\nu-2\chi^2_{mode}=\nu-2$$, while $$E[\chi^2]=\nuE[\chi^2]=\nu$$, where $$\nu\nu$$ are the degrees of freedom of data.

I heard at university that, when doing a $$\chi^2\chi^2$$ test after using Least Squares Method to fit a set of data, I should expect to get $$\chi^2 \approx \nu\chi^2 \approx \nu$$ because “that’s what happens in general”.

Did I misunderstand all of this or is the expected value somehow very probable? (Even if the most probable value is of course the mode)