Find random variable $X$ with $\text{E}(X) = 0$, $\text{Var}(X) = 4$, and $\text{P}( \vert X \vert \geq 4 ) = 0.25$ As the title states, my task is to find a random variable $X$ with $\text{E}(X) = 0$, $\text{Var}(X) = 4$, and $\text{P}( \vert X \vert \geq 4 ) = 0.25$.
My first attempt was to consider $X \sim \text{Normal}(0,2^2)$, but with this we have
$$ \text{P}( \vert X \vert \geq 4 ) = 1 - \text{P}(-4 < X < 4) = 1 - \text{P}(-2 < Z < 2) \approx 0.0455 < 0.25  $$
and so this choice of $X$ is obviously incorrect.
My second thought was to try $X \sim \text{Uniform}(a,b)$ such that $\text{E}(X) = \frac{a+b}{2} = 0$, which is to say $a = -b$, and $\text{Var}(X) = \frac{(a-b)^2}{12} = \frac{b^2}{3} = 4$, which to say $b = \sqrt{12} \approx 3.4641$. However, using this we then have
$$ \text{P}( \vert X \vert \geq 4 ) = 1 - \text{P}(-4 < X < 4) = 1 - 1 = 0 $$
and so this choice of $X$ must also be incorrect.
With the provided information, it must be true that $X$ can take both positive and negative values and I think it must be true that $X$ should be absolutely continuous (this may be incorrect, however). With these thoughts in mind and letting $f_X(x)$ represent the PDF of X for $x \in A \subseteq \mathbb{R}$, I know $X$ must satisfy the following:
$(i)$ $\text{  } \text{E}(X) = \int_{\mathbb{R}} x f_X(x) \text{d}x = 0$
$(ii)$ $\text{  } \text{Var}(X) = \text{E}(X^2) = \int_{\mathbb{R}} x^2 f_X(x) \text{d}x = 4$
$(iii)$ $\text{  } \int_{-4}^{4} f_X(x) \text{d}x = 0.25 $
I am unsure as to how to go about determining $f_X(x)$ under these conditions. I was hoping the distribution of $X$ would be something quite simple but it may be more complicated than I was thinking, perhaps $X$ is even a mix of absolutely continuous and discrete parts. Any direction would be greatly appreciated.
 A: Consider the discrete random variable $X$ with PMF given by
$$\text{P}(X = x) = \begin{cases}
\frac{3}{4} & \text{ for } x = 0 \\
0 & \text{ for } \vert x \vert = 1 \\
0 & \text{ for } \vert x \vert = 2 \\
0 & \text{ for } \vert x \vert = 3 \\
\frac{1}{8} & \text{ for } \vert x \vert = 4 \\
0 & \text{ for all other $x \in \mathbb{Z}$ }
\end{cases}$$
With this random variable, it is clear that
$$ \sum_{k = -4}^{4} \text{P}(X = k) = \frac{1}{8} + \frac{3}{4} + \frac{1}{8} = 1 $$
and the desired conditions are satisfied:
$$ \text{E}(X) = \sum_{k = -4}^{4} k \text{P}(X = k) = (-4)\frac{1}{8} + (4)\frac{1}{8} = 0 $$
$$ \text{Var}(X) = \sum_{k = -4}^{4} k^2 \text{P}(X = k) =(16)\frac{1}{8} + (16)\frac{1}{8} = 4 $$
$$ \text{P}(\vert X \vert \geq 4) = \text{P}(X = -4) + \text{P}(X = 4) = \frac{1}{4} $$
A: As the comment suggested, it would be way easier to construct a simple discrete example instead of a continuous one.  In other words, you should consider rewriting equations (i) -- (iii) in your question body in terms of summations.  Furthermore, the given information clearly suggested somewhat symmetry, which you should also take advantage of.
One possible answer:  Let $X$ be a discrete (three-point distribution) $X \in \{-4, 0, 4\}$ with probability mass function $P(X = 4) = P(X = -4) = 1/8$, and $P(X = 0) = 3/4$.
More interestingly, below is an analytical derivation of this is actually the unique random variable satisfying given conditions, as @Daniel, StubbornAtom hinted.
Since
\begin{align}
& E(X^2) = E(X^2I_{\{|X| < 4\}}) + E(X^2I_{\{|X| \geq 4\}}) \geq  E(X^2I_{\{|X| < 4\}}) + 16P(|X| \geq 4), \\
& E(X^2) = 16P(|X| \geq 4) = 4, 
\end{align}
it follows that $E(X^2I_{\{|X| < 4\}}) = 0$, which requires $P[X^2I_{\{|X| < 4\}} > 0] = P[0 < |X| < 4] = 0$.
On the other hand, $E((X^2 - 16)I_{\{|X| \geq 4\}}) = 0$ implies
\begin{align}
  & P[(X^2 - 16)I_{\{|X| \geq 4\}} > 0] \\
= & P[(X^2 - 16)I_{\{|X| \geq 4\}} > 0, |X| = 4] + 
    P[(X^2 - 16)I_{\{|X| \geq 4\}} > 0, |X| > 4] \\
= & P[|X| > 4] = 0.  
\end{align}
In the above, we used the fact that if a non-negative random variable $Y$ satisfies $E(Y) = 0$, then $P(Y > 0) = 0$ (Probability and Measure, Theorem $15.2$(ii)).
In summary, $X$ can only have probability masses at $\{0, 4, -4\}$, which then entails $P(X = 4) = P(X = -4) = 1/8$, and $P(X = 0) = 3/4$ by conditions $E(X) = 0$ and $E(X^2) = 4$.
