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PhD Student in ECE @ MSU

Find the Expectation of a Symmetric Probabiliry Density Function


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.


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

\[\begin{align} E(X)&=\int_{-\infty}^{\infty}xf(x)dx\\ &=\int_{-\infty}^{\infty}(c+(x-c))f(x)dx \\ &= \int_{-\infty}^{\infty}cf(x)dx + \int_{-\infty}^{\infty}(x-c)f(x)dx \\ &= c + \int_{-\infty}^{\infty}(x-c)f(x)dx \end{align}\]

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

\[\begin{align} \int_{-\infty}^{\infty}(x-c)f(x)dx &= \int_{-\infty}^{\infty}hf(h+c)dh\\ &=\int_{-\infty}^{0}hf(h+c)dh+\int_{0}^{\infty}hf(h+c)dh \\ &= -\int_{0}^{\infty}(-m)f(c-m)d(-m) +\int_{0}^{\infty}hf(h+c)dh \\ &= -\int_{0}^{\infty}mf(c-m)dm +\int_{0}^{\infty}hf(h+c)dh \\ &= 0 \end{align}\]


  1. Proof of $E(X)=aE(X)=a$ when $a$ is a point of symmetry 

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