Show that the distribution of $X$ is like this:

Question

Suppose $E(X)=\mu$ and $V(X)=\sigma^2$. Show that if $P(|X-\mu|\geq k)=\frac{\sigma^2}{k^2}$ then we have: $$P(X=\mu)=\frac{k^2-\sigma^2}{k^2}, P(X=\mu-k)=P(X=\mu+k)=\frac{\sigma^2}{2k^2}$$

I know how to prove the inverse of this theorem. But I have no idea how to prove this side. Would you help me? Thank you in advance.

Answer 1

Assume $0<\sigma\leq k$. First, we show $\text{supp}\{X\}\subseteq\{\mu,\mu+k,\mu-k\}.$

Lemma: Suppose $Z\geq 0,a>0$ and $P(Z\geq a)=E[Z]/a.$ Then $\text{supp}\{Z\}\subseteq\{0,a\}$

Proof: Note ${\bf 1}_{Z\geq a}\leq Z/a$ for $Z\geq 0$ with equality only holding at $Z=0,a.$ Hence $P(Z\geq a)=E[{\bf 1}_{Z\geq a}]= E[Z/a]$ only if $Z=0$ or $Z=a$ a.s.$\square$

Now we are given $\frac{1}{k^2/\sigma^2}=P(|X-\mu|\geq k)=P\left(\frac{(X-\mu)^2}{\sigma^2}\geq \frac{k^2}{\sigma^2}\right),$ so the claimed three-point support follows immediately from our lemma.

Now we simply need to find the mass on each point. Note $$P(X=\mu)=P(|X-\mu|< k)=1-P(|X-\mu|\geq k)=1-\frac{\sigma^2}{k^2}.$$

Noting the three masses sum to unity, and $\mu+k$ and $\mu-k$ are equidistant from mean $\mu$, the other masses follow:

$$P(X=\mu+k)=P(X=\mu-k)=\frac{\sigma^2}{2k^2}.$$

Answer 2

Now in the following I will assume that $X$ is discrete and that it has as support all the integers $\mathbb{Z}$. Note that from $\mathbb{P}(\lvert X-\mu\rvert\geq k) = \frac{\sigma^2}{k^ 2}$,

\begin{align} \mathbb{P}(\lvert X-\mu\rvert < k) = 1 - \frac{\sigma^2}{k^ 2} = \frac{k^ 2 - \sigma^2}{k^ 2}. \end{align} Which implies, \begin{align} \mathbb{P}(X = \mu) = \mathbb{P}(\lvert X-\mu\rvert = 0) = \mathbb{P}(\lvert X-\mu\rvert < 1) = 1 - \sigma^2. \end{align}

Note, \begin{align} \mathbb{P}(\lvert X-\mu\rvert \geq k) &= \mathbb{P}(\{X-\mu\geq k\}\cup\{X-\mu \leq -k\}), \\[1em] &\stackrel{m.e.}{=} \mathbb{P}( X \geq \mu + k) + \mathbb{P}( X \leq \mu - k) = \frac{\sigma^2}{k^2}. \end{align} So that, \begin{align} \frac{\sigma^2}{k^2} - \frac{\sigma^2}{(k+1)^2} &= \mathbb{P}( X \geq \mu + k) + \mathbb{P}( X \leq \mu - k) - \left(\mathbb{P}( X \geq \mu + k + 1) + \mathbb{P}( X \leq \mu - k-1)\right), \\[1em] &= \mathbb{P}( X \geq \mu + k) - \mathbb{P}( X \geq \mu + k + 1) + \mathbb{P}( X \leq \mu - k) - \mathbb{P}( X \leq \mu - k-1), \\[1em] &= \mathbb{P}(X=\mu+k) + \mathbb{P}(X=\mu - k). \end{align}

Assuming that the support equals the integers, we must have, \begin{align} \mathbb{P}(X=\mu)+\sum_{k=1}^{\infty} \mathbb{P}(X=\mu+k) + \mathbb{P}(X=\mu - k) &= 1-\sigma^2 + \sigma^2\sum_{k=1}^{\infty} \left(\frac{1}{k^2} - \frac{1}{(k+1)^2}\right) \\[1em] &= 1-\sigma^2 +\sigma^2\left(\left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \dots\right) \\[1em] &= 1. \end{align}

Nevertheless, since,

$$\mathbb{P}(X=\mu+k) + \mathbb{P}(X=\mu - k) = \sigma^2\left(\frac{1}{k^2} - \frac{1}{(k+1)^2}\right).$$

These probabilities cannot equal each other and be equal to $\frac{\sigma^2}{2k}$ as you suggest.