probability: When $X\sim\mathcal{N}(\mu, \sigma^2)$, the med = $\mu$

Question

The median of a PDF is defined as the point $x = $med for which $P[X\leq\text{med}] = 1/2$. Prove that if $X\sim\mathcal{N}(\mu, \sigma^2)$, then med $= \mu$.

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Since $P[X\leq\text{med}] = 1/2$, we need to determine $$ \Phi(x) - \Phi(-\infty) = 1 - Q(x) - 1 - Q(-\infty) = 1 - Q(x) = \frac{1}{2} $$ Thus, the we need the value of $x$ such that $Q(x) = 1/2$. \begin{gather*} Q(x) = \frac{1}{\sqrt{2\pi}}\int_{\frac{x - \mu}{\sigma}}^{\infty}e^{-t^2/2}dt\\ \frac{1}{2}\text{erfc}\Big[\frac{x - \mu}{\sigma\sqrt{2}}\Big] = \frac{1}{2}\\ \text{erfc}\Big[\frac{x - \mu}{\sigma\sqrt{2}}\Big] = 1 \end{gather*} I know that $\text{erfc}(0) = 1$ but is there a more succinct way to prove $x = \mu$ without saying from prior knowledge we know that $\text{erfc}(0) = 1$ so $x$ must be $\mu$?

Answer 1

The below theorem will make our proof very easy.

If $X\sim N(\mu,\sigma^2)$, then pdf is symmetric about $\mu$

Proof: Note that proving symmetry about $\mu$ boils down to proving that for $\epsilon>0$ that $f(\mu+\epsilon)=f(\mu-\epsilon)$. thus we see that $$f(\mu-\epsilon)=\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{\left((\mu-\epsilon)-\mu\right)^2}{\sigma^2}}=\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{\left(-\epsilon\right)^2}{\sigma^2}}=\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{\left(\epsilon\right)^2}{\sigma^2}}$$ and $$f(\mu+\epsilon)=\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{\left((\mu+\epsilon)-\mu\right)^2}{\sigma^2}}=\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{\left(\epsilon\right)^2}{\sigma^2}}$$ thus $f(\mu+\epsilon)=f(\mu-\epsilon)$.

With this theorem we now see that $$P(X>\mu)=\int_\mu^\infty\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{\left(x-\mu\right)^2}{\sigma^2}}dx=P(X<\mu)$$ And since $P(X>\mu)+P(X<\mu)=1$ we have $2P(X>\mu)=1$ thus $P(X>\mu)=\frac{1}{2}$

Answer 2

The normal pdf is symmetric about the mean, isn't that enough?