2
$\begingroup$

I'm struggling to prove a very simple intuition. Adding more of the largest observed value in a mean will increase the mean's value.

For example $X = \{1,2,3\} \implies \mu = 2$;$X = \{1,2,3,3\} \implies \mu = 2.25$.

This intuition should be true even for a continuous probability distribution. Imagine the graph of a p.d.f. of a normal distribution. If one "pulls up" the right side of the distribution, the mean should move to the right.

My question is: How does one prove that for a continuous p.d.f.?

I can prove that on a discrete distribution relatively easily:

$$ \mu = \sum_i w_ix_i \implies \frac{\partial \mu}{\partial w_{i|max\, x}}= x_{i|max\, x}>0$$

In English, the average of a discrete p.d.f. is just a weighted sum where the weights sum to 1 (these weights represent the probability of observing this value). If I take a derivative with respect to the weight of the largest value, it is easy to see that this derivative is positive. In other words, increasing the weight of the maximum value increases the mean overall.

However, this proof doesn't quite work for a continuous p.d.f. $max\, x$ can be infinite, and $p(x)$ in a single value tends to 0. Plus trying to integrate by parts of the law of unconscious statistician gives seems to cancel out the probability. Any ideas how can I prove such a simple intuition? Bonus points if we can show the same relation for skewness.

$\endgroup$
1
  • 2
    $\begingroup$ Can't you represent this as a new random variable $y = \lambda x + (1-\lambda) x_{max}$? In that case, $Ey = \lambda Ex + (1-\lambda) x_{max} \ge Ex$. $\endgroup$ Aug 2, 2023 at 13:25

1 Answer 1

2
$\begingroup$

If you consider the "balance point" of a continuous function f(x)

$$\mu_1=\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}$$

$\mu_1$ can be increased by adding $g(x)$ such that $$\begin{align}\int_{\mu_1}^bg(x)dx>\int_a^{\mu_1}g(x)dx\Leftrightarrow\mu_1&<\frac{\int_a^bxg(x)dx}{\int_a^bg(x)dx}\\ &=\frac{\int_a^bxg(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}\frac{\int_a^b\left(f(x)+g(x)\right)dx}{\int_a^bg(x)dx} \end{align}$$ $$ \begin{align} \therefore\mu_1\frac{\int_a^bg(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}-\mu_1&<\frac{\int_a^bxg(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}-\mu_1\\ \mu_1\frac{\int_a^bg(x)dx-\int_a^b\left(f(x)+g(x)\right)dx}{\int_a^b\left(f(x)+g(x)\right)dx}&<\frac{\int_a^bxg(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}-\mu_1\\ -\mu_1\frac{\int_a^bf(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}&<\frac{\int_a^bxg(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}-\mu_1\\ \mu_1&<\frac{\int_a^bxg(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}+\mu_1\frac{\int_a^b f(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}\\ &=\frac{\int_a^bxg(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}+\frac{\int_a^b xf(x)dx}{\int_a^b\left(f(x)+g(x)\right)dx}\\ &=\frac{\int_a^bx\left(f(x)+g(x)\right)dx}{\int_a^b\left(f(x)+g(x)\right)dx}\\ &=\mu_2 \end{align} $$

$\endgroup$
5
  • $\begingroup$ Thanks a lot! Just one detail, I'm missing: how did you got $\int_a^b f(x)dx$ in the numerator of the second term? I assume its something with $-\mu_1$ but cant figure it out $\endgroup$
    – JMenezes
    Aug 2, 2023 at 17:43
  • $\begingroup$ Just subtract $\int_a^b(f(x)+g(x))dx$ from $\int_a^bg(x)dx$ $\endgroup$
    – Localth
    Aug 2, 2023 at 17:49
  • $\begingroup$ @JMenezes Did that make sense of it for you? $\endgroup$
    – Localth
    Aug 2, 2023 at 18:23
  • $\begingroup$ Sorry, I tried a couple more hours, and called another friend but we still don't see it. I understand you can transform the left-hand side in $\mu_1 \frac{\int_a^b f(x) dx}{\int_a^b f(x) +g(x) dx}$, but shouldn't that need an inversion to be moved to the other side, placing the $\int_a^b f(x) dx$ in the denominator $\endgroup$
    – JMenezes
    Aug 3, 2023 at 8:55
  • $\begingroup$ I've interposed two lines so you should see it now $\endgroup$
    – Localth
    Aug 3, 2023 at 11:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .