Does $E(X-E(X))=0$? - Mathematics Stack Exchange

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Learn more about Teams Does $E(X-E(X))=0$? Ask Question Asked 10 years, 3 months ago Modified 10 years, 3 months ago Viewed 19k times 10 $\begingroup$

In a book I was reading, it seemed to imply that $E(X-E(X))=0$. My intuition tells me this is true, because if $E(X)$ is the "centre", then the average displacement from this centre should be 0. However, can someone show me a formal proof (assuming it is true)?

• expectation

Share Cite Follow asked Sep 5, 2014 at 20:37 someguysomeguy 24311 gold badge33 silver badges99 bronze badges $\endgroup$ 1

• 3 $\begingroup$ Yes, use aditivitty of the expected value and E(E(X)) = E(X) $\endgroup$ – PenasRaul Commented Sep 5, 2014 at 20:38

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5 Answers 5

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 17 $\begingroup$

It is true by the linearity of expectation:

$$E(X - E(X)) = E(X) - E(E(X))$$

Since $E(X)$ is a constant $E(E(X))$ is just $E(X)$ and therefore $E(X) - E(E(X)) = 0$

Share Cite Follow answered Sep 5, 2014 at 20:40 NullNull 1,35233 gold badges1818 silver badges2222 bronze badges $\endgroup$ 1

• $\begingroup$ Thank you. What confused me when I was trying to derive this is E(E(X)). In my mind I was thinking E(X) is a function of X, but of course it's a constant. $\endgroup$ – someguy Commented Sep 5, 2014 at 20:43

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Expectations are additive: $$E(A+B) = E(A) + E(B)$$

So in particular, $$\begin{align}E(X-E(X)) &= E(X) - E(E(X))\\ & = E(X) - E(X)\\ &= 0.\end{align}$$ where the second line follows because $E(X)$ is a constant and $E(C) = C$ for constants.

Share Cite Follow answered Sep 5, 2014 at 20:39 community wiki MJD $\endgroup$ Add a comment | 2 $\begingroup$

$$ \mathbb{E}(X-\mathbb{E}(X)) = \mathbb{E}(X) - \mathbb{E}(\mathbb{E}(X)) = \mathbb{E}(X) - \mathbb{E}(X) = 0.$$

Or even $$ \mathbb{E}(X-\mathbb{E}(X))^2 = \mathbb{E}((X-\mathbb{E}(X)^2) - \mbox{Var}(X-\mathbb{E}(X)) = \mbox{Var}(X) - \mbox{Var}(X) = 0. $$

Share Cite Follow edited Sep 5, 2014 at 20:55 answered Sep 5, 2014 at 20:40 user133281user133281 16.2k22 gold badges3737 silver badges6464 bronze badges $\endgroup$ 1

• $\begingroup$ The second option requires the assumption that $E[X^2]< \infty$, while the statement that $E[X-E[X]] = 0$ only requires $E[X]$ exist (i.e. $E[|X|] < \infty$) [$E[X^2]<\infty$ obviously implies $E[X]$ exists]. So if I were grading the question, I'd probably dock points for the second answer. $\endgroup$ – Batman Commented Sep 5, 2014 at 21:10

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Yes, it's true in general by linearity of expectations.

But as an illustration, consider the discrete, finite case with $n$ values: $X=\{x_1,x_2,\dots,x_n\}$ each occurring with equal probability.

Then

$$ E(X)=\dfrac{1}{n} \sum_{i=1}^{n} x_i $$

and

$$ E(X-E(X))=\dfrac{1}{n}\sum_{j=1}^{n}\left(x_j-\dfrac{1}{n}\sum_{i=1}^{n}x_i \right)=\dfrac{1}{n}\left(\sum_{j=1}^{n}x_j - \sum_{i=1}^n x_i\right)=0 $$

In the last step we have moved the sum over $i$ outside the sum over $j$, because the former does not depend on $j$. We need to multiply the $\sum_i$ term by $n$, however, as $\sum_j$ has $n$ terms. This $n$ cancels out with the $1/n$ associated with the $\sum_i$ term.

Share Cite Follow edited Sep 6, 2014 at 4:36 answered Sep 5, 2014 at 20:41 MGAMGA 9,67644 gold badges4242 silver badges5959 bronze badges $\endgroup$ 6

• $\begingroup$ You implicitly assume each x has equal probability of being drawn. $\endgroup$ – candido Commented Sep 5, 2014 at 21:10
• $\begingroup$ @candido I'm just talking about the average of a set of numbers. $\endgroup$ – MGA Commented Sep 5, 2014 at 21:56
• $\begingroup$ Right, the average of a a set of numbers is as you defined it, but the expected value is not necessarily an average that gives each value the same weight. E(X) is not quite how you defined it. You are correct that E(X) is how you defined it ONLY IF each value of x has an equal probability of being drawn, since then Pr(xi)=1/n like you defined it. In other words, the way you used E(X) is a special case of the more general definition of E(X) with a general density f(x). en.wikipedia.org/wiki/Expected_value $\endgroup$ – candido Commented Sep 6, 2014 at 3:25
• $\begingroup$ @candido Yes yes, I know, and I agree with everything you said. I just wanted to give the OP an illustration for the simplest possible case, that's all :) $\endgroup$ – MGA Commented Sep 6, 2014 at 4:31
• $\begingroup$ Ah ok! Didn't mean to be picky, just thought I'd help just in case :) $\endgroup$ – candido Commented Sep 6, 2014 at 4:32

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More Generally, note that for a random variable (r.v.) X can have a discrete distribution with probabilities that X can take a certain value x is P(X=x) (the probability mass function), a continuous distribution, where X has a continuous probability density function f(x) (or a mixture for a set of domains). If X is wholly discrete f(x) is zero for all x, likewise a fully continuous r.v. means P(X=x) = 0 for all x.

$$E(X) = \sum_{x=0}^\infty xP(X=x) + \int_0^\infty xf(x) \;\mathrm{dx}$$

Which means

$$E(X-E(X)) = \sum_{x=0}^\infty \left(x- \left[\sum_{x=0}^\infty xP(X=x) + \int_0^\infty xf(x) \;\mathrm{dx}\right]\right)P(X=x) + \int_0^\infty \left(x-\left[\sum_{x=0}^\infty xP(X=x) + \int_0^\infty xf(x) \;\mathrm{dx}\right]\right)f(x) \;\mathrm{dx}$$

This may seems like overkill BUT you can then just follow the algebra without quoting presumed theorems (linearity of Expectation etc) - which can easily be derived anyway...

Share Cite Follow edited Sep 5, 2014 at 22:53 Michael Hardy 1 answered Sep 5, 2014 at 21:34 WeibullWeibull 1133 bronze badges $\endgroup$ 3

• $\begingroup$ Why do you exclude "$+$" and "$=$" from your MathJax code?? (I corrected it.) $\endgroup$ – Michael Hardy Commented Sep 5, 2014 at 22:55
• $\begingroup$ Thanks, I'm new to the code; didn't realise it made a difference haha $\endgroup$ – Weibull Commented Sep 6, 2014 at 12:05
• $\begingroup$ The difference is huge if you do it with a minus sign: $\displaystyle\int f(x)\,dx-\int g(x)\,dx$ versus $\displaystyle\int f(x)\,dx$-$\int g(x)\,dx$ ${}\qquad{}$ $\endgroup$ – Michael Hardy Commented Sep 6, 2014 at 16:11

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