I am trying to develop a proof of the following:

Given a random variable $X$ with symmetric probability density function $f(x)$, prove that $E(X)=a$ where $a$ is the point of symmetry.

A couple of thoughts:

It's easy to think of examples where this applies (e.g. normal) and doesn't (e.g. Cauchy). I am comfortable with calculating expectation, and even handling showing how


But if the specific form of the pdf is not specified, is it possible to prove this in general?

Also, how would one go about showing the existence of the expectation in general?

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    $\begingroup$ The result is not true, the expectation $E(X)$ may not exist. But if expectation does exist, the result is true. $\endgroup$ Commented Sep 10, 2013 at 1:32
  • $\begingroup$ @AndréNicolas, I agree but is there a way to prove that $E(X)$ exists in general, or must we know the form of the pdf or cdf to proceed? $\endgroup$
    – Justin
    Commented Sep 10, 2013 at 1:35

1 Answer 1


We give two proofs, the first by manipulation of the integral, and a second much shorter one that uses probabilistic language.

Proof 1: Symmetry about $a$ means that $f(a+z)=f(a-z)$ for all $z$.

Suppose that the expectation $E(X)$ exists. Then $$E(X)=\int_{-\infty}^\infty xf(x)\,dx.$$ Rewrite as $$\int_{-\infty}^\infty \left(a +(x-a)\right)f(x)\,dx,$$ and then as $$\int_{-\infty}^\infty \left(a +(x-a)\right)f(x)\,dx,$$ which is equal to $$\int_{-\infty}^\infty af(x)\,dx+\int_{-\infty}^\infty (x-a)f(x)\,dx.\tag{1}$$ The first integral is $a$. We need to show the second integral is $0$. This is essentially obvious by symmetry: $(x-a)f(x)$ is an odd function. But we do the details.

In the second integral of (1), make the change of variable $y=x-a$. Then the integral becomes $$\int_{-\infty}^\infty yf(a+y)\,dy.$$ Break up at $y=0$. We get $$\int_{-\infty}^0 yf(a+y)\,dy +\int_0^\infty yf(a+y)\,dy.\tag{2}$$ For the first integral, make the change of variable $z=-y$. We get $$\int_{z=\infty}^0 (-1)(-z)f(a-z)\,dz.$$ Using $f(a-z)=f(a+z)$, and some minor fooling with minus signs, we end up with $$-\int_{z=0}^\infty zf(a+z)\,dz.$$ This cancels the second integral of (2).

Proof 2: A more probabilistic (and therefore better) way of doing the problem is to first let $X=a+Y$. Assume that $E(X)$ exists (it need not). Then $E(X)=a+E(Y)$. We need to show that $E(Y)=0$.

The density function $g(y)$ of $Y$ is symmetric about $y=0$. It follows that that the random variable $Y$ has the same distribution as the random variable $-Y$. Thus $E(Y)=E(-Y)=-E(Y)$ and therefore each is $0$.

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    $\begingroup$ Excellent and clear answer. Thank you for that. The only place I am not clear is why the second to last line is z=$\infty$. I notice that you begin by assuming that the expectation exists. Is it even possible to show/prove the existence in general? I assume not since there are obvious examples of symmetric pdfs both with and without finite expectation $\endgroup$
    – Justin
    Commented Sep 10, 2013 at 2:05
  • $\begingroup$ I have added a better answer, more conceptual, and much easier to type! The first one was correct but "crude." $\endgroup$ Commented Sep 10, 2013 at 2:08
  • $\begingroup$ Thank you again. The second approach is not one which I have seen before, but it is simple and elegant. $\endgroup$
    – Justin
    Commented Sep 10, 2013 at 2:09
  • $\begingroup$ In Proof2, perhaps it is necessary to say in the last sentence that $E[Y]$ if it exists must be the same as $E[-Y] = -E[Y]$? Else the argument would serve to show that a standard Cauchy random variable $Y$ has zero mean which is not so. $\endgroup$ Commented Sep 10, 2013 at 2:35
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    $\begingroup$ You are right, I stressed it it exists in the first proof, and should repeat in the second proof, which may be read separately. $\endgroup$ Commented Sep 10, 2013 at 2:38

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