What is mean and median of $X$ if $P(X \geq \frac{1}{2}+x)=P(X \leq \frac{1}{2}-x)$?

Question

Let $X$ be a random variable such that $E[X] < \infty$ and $P(X \geq \frac{1}{2}+x)=P(X \leq \frac{1}{2}-x), \forall x \in R$ then find $E(X)$ and median$(X)$.

My attempt:

$\int_{-\infty}^{\frac{1}{2}-x} f(t) dt = \int_{\frac{1}{2}+x}^{+\infty} f(t) dt \\ E[X] = \int_{-\infty}^{+\infty} x f(x) dx $

Now i tried splitting interval into $(-\infty,\frac{1}{2}-x], [\frac{1}{2}-x, \frac{1}{2}+x] , [\frac{1}{2}+x, +\infty)$ and intergrate by parts. But i am not to proceed. Kindly tell me how to proceed

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Answer 1

The fact that: $$P\left(X\geq \frac{1}{2}+x\right)=P\left(X\leq \frac{1}{2}-x\right)\text{ for all }x\in\mathbb R$$tells us that:$$X-\frac{1}{2}\stackrel{d}{=}\frac{1}{2}-X$$ so that - if $\mathbb{E}X$ exists - we have: $$\mathbb{E}X-\frac{1}{2}=\mathbb{E}\left(X-\frac{1}{2}\right)=\mathbb{E}\left(\frac{1}{2}-X\right)=\frac{1}{2}-\mathbb{E}X$$ or equivalently: $$\mathbb{E}X=\frac{1}{2}$$

$m$ is a median iff: $$P\left(X\geq m\right)\geq\frac{1}{2}\text{ and }P\left(X\leq m\right)\geq\frac{1}{2}$$ Observe that: $$P\left(X\geq \frac{1}{2}\right)=P\left(X\leq \frac{1}{2}\right)$$ and also: $$P\left(X\geq \frac{1}{2}\right)+P\left(X\leq \frac{1}{2}\right)=1+P\left(X=\frac12\right)\geq1$$ implying that: $$P\left(X\geq \frac{1}{2}\right)\geq\frac{1}{2}\text{ and }P\left(X\leq \frac{1}{2}\right)\geq\frac{1}{2}$$ This proves that $\frac{1}{2}$ is a median.

It is not excluded that there are more medians though.

Answer 2

The condition, $\forall x \in \mathbb{R}$

$$\mathbb{P}[X \leq \frac{1}{2}-x ]=\mathbb{P}[X \geq \frac{1}{2}+x ]$$

implies that the density of X is symmetric around $\frac{1}{2}$ and unimodal.

Thus Mean, Mode and Median is $\frac{1}{2}$

Answer 3

Let, $T_1 = X-\frac{1}{2}$

Now consider the another form of expectation using CDF function,

$$\mathbb{E}(T_1)= \int_0^{\infty}\big[\mathbb{P}(T_1>x)- \mathbb{P}(T_1\leq-x)\big]dx\\ =\int_0^{\infty}\big[\mathbb{P}(X>\frac{1}{2}+x)- \mathbb{P}(X\leq\frac{1}{2}-x)\big]dx\\ =0\\ \Rightarrow \mathbb{E}\big(X-\frac{1}{2}\big)=0\Rightarrow \mathbb{E}(X)=\frac{1}{2}$$

And for getting median $=\frac{1}{2}$, previous answers are enough.