<s>[INST] 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.
 [/INST] 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:
$$ \text{E}(X) = \sum_{k = -4}^{4} k \text{P}(X = k) = (-4)\frac{1}{8} + (4)\frac{1}{8} = 0 $$
$$ \text{Var}(X) = \sum_{k = -4}^{4} k^2 \text{P}(X = k) =(16)\frac{1}{8} + (16)\frac{1}{8} = 4 $$
$$ \text{P}(\vert X \vert \geq 4) = \text{P}(X = -4) + \text{P}(X = 4) = \frac{1}{4} $$
</s>