# &#x03BC;<!-- μ --> = m = q 1 </msub>

$\mu =m=\frac{{q}_{1}+{q}_{3}}{2}$
Let $X$ be a random variable with p.m.f./p.d.f. ${f}_{X}\left(x\right)$ that is symmetric about $\mu \left(\in R\right),$ i.e., ${f}_{X}\left(x+\mu \right)={f}_{X}\left(\mu -x\right)$, $\mathrm{\forall }x\in \left(-\mathrm{\infty },\mathrm{\infty }\right).$
If ${q}_{1},m$ and ${q}_{3}$ are respectively the lower quartile, the median and the upper quartile of the distribution of $X$ then show that $\mu =m=\frac{{q}_{1}+{q}_{3}}{2}$
How to prove.....
any Hints.
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Mackenzie Zimmerman
Say, ${q}_{1}$ is the lower quartile, i.e. $P\left(X<{q}_{1}\right)=1/4$, that is, ${\int }_{-\mathrm{\infty }}^{{q}_{1}}{f}_{X}\left(x\right)dx=1/4$. By symmetry about $\mu$, (by ), we also have ${\int }_{2\mu -{q}_{1}}^{\mathrm{\infty }}{f}_{X}\left(x\right)dx=1/4$, that is, $P\left(X>2\mu -{q}_{1}\right)=1/4$, so ${q}_{3}=2\mu -{q}_{1}$
Similarly, $P\left(X<\mu \right)=1/2$ by symmetry.
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Marissa Singh
Just want to add another interesting tidbit - if $E\left[X\right]<\mathrm{\infty }$, then
$E\left[X\right]=\mu =m=\frac{{q}_{1}+{q}_{3}}{2}$
$\begin{array}{rl}E\left[X\right]& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{f}_{X}\left(x\right)dx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left(2\mu -x\right){f}_{X}\left(x\right)dx\\ & =2\mu {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}_{X}\left(x\right)dx-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{f}_{X}\left(x\right)dx\\ & =2\mu -E\left[X\right]\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}E\left[X\right]& =\mu \end{array}$