Computing mean of probability density function without integration

Question

Is there any method of determining the mean of a random variable $X$ without integration?

hey, yep I've been looking at the graph for a while, its symmetrical but I'm not sure how that helps me exactly — , Apr 26 '16 at 12:34

score 1 · Answer 1 · answered Apr 26 '16 at 13:09

1

As $$f_X(1+\delta) = f_X(1-\delta)$$ for $\delta>0$ the mean and the median are the same. Therefore $\mathbb{E}[X]=1$.

answered Apr 26 '16 at 13:09

I don't quite understand exactly what you mean, specifically I'm unfamiliar with the notation. Is there a simpler explanation you can give me? thanks. – Apr 26 '16 at 13:32
The function is symmetrical in relation to 1. Therefore, every marginal gain has an opposite loss, which leads the mean to be 1. – Guilherme Thompson Apr 26 '16 at 13:36

score 1 · Answer 2 · answered Apr 26 '16 at 13:37

You will have two symmetrical traingles mirrored at x= 1. The mean of this pdf in terms of integration is

$E(k) = \int_{0}^{1} k(1-k)dk + \int_{1}^{2}k(k-1)dk = 1$

Just what the other answers have indicated the Mean is simply 1 by symmetry.

The mean for the first triangle is the k which splits the area of the triangle into two halves. That $k_1 = \frac{3-\sqrt{5}}{2}$

The mean for the second triangle is the k which splits the area of the triangle into two halves. That $k_2 = \frac{1+\sqrt{5}}{2}$.

Thus the mean of the pdf $= \frac{k_1+k_2}{2} = 1$

score 0 · Answer 3 · edited Apr 13 '17 at 12:19

0

The mean of a symmetric pdf, if it exists, is at the point of symmetry. See here. Furthermore, if the support is bounded then the mean exists.

edited Apr 13 '17 at 12:19

Community

answered Apr 26 '16 at 12:42

leonbloy

3 Answers3