Find the Expectation of a Symmetric Probabiliry Density Function

2017-10-18

If the PDF is symmetric about c, show that E(X)=c. This is a homework problem for course STT802-002 Theory of Probabilities and Statistics I in MSU.

Problems

Suppose that X has a density $f$ that is symmetric about $c$ . That is, $f (c + h) = f (c - h)$ for all real $h$ . Show that, if it exists, $E (X) = c$ . Hint: Make the change of variable $h = x - c$ .

Solution ¹

\begin{aligned} E (X) & = \int_{- \infty}^{\infty} x f (x) d x \\ = \int_{- \infty}^{\infty} (c + (x - c)) f (x) d x \\ = \int_{- \infty}^{\infty} c f (x) d x + \int_{- \infty}^{\infty} (x - c) f (x) d x \\ = c + \int_{- \infty}^{\infty} (x - c) f (x) d x \end{aligned}

We have already had $c$ in the expression, we just need to prove the second term is 0. Let $h = x - c$ , then $d h = d x$ , $x = h + c$ .

\begin{aligned} \int_{- \infty}^{\infty} (x - c) f (x) d x & = \int_{- \infty}^{\infty} h f (h + c) d h \\ = \int_{- \infty}^{0} h f (h + c) d h + \int_{0}^{\infty} h f (h + c) d h \\ = - \int_{0}^{\infty} (- m) f (c - m) d (- m) + \int_{0}^{\infty} h f (h + c) d h \\ = - \int_{0}^{\infty} m f (c - m) d m + \int_{0}^{\infty} h f (h + c) d h \\ = 0 \end{aligned}

Therefore,

E (X) = c + 0 = c

Proof of $E (X) = a E (X) = a$ when $a$ is a point of symmetry ↩

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Find the Expectation of a Symmetric Probabiliry Density Function

Problems

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Solution 1

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