i'm stuck with this problem.Can anyone please give me a help?
"Find the distribution of a random variable $\mathcal{X}$ such that suppose has the following properties
$ m=\mathbb{P( \mathcal{X} = 1})= 1 - \mathbb{P}(\mathcal{X}=-1)$ and $ \mathbb{E} (\mathcal{X}) = Var ( \mathcal{X}) $"
My reasoning is $ \mathbb{P}(\mathcal{X}=-1)= 1-m$ then $\mathbb{E} (\mathcal{X}) = 2m-1 $= $Var ( \mathcal{X})= 1-(4m^2 -4m+1)=-4m^2 +4m $
This implies
$$4m^2 -2m+1=0$$ and the positive root of this polynomial is $ \mathbb{P}( \mathcal{X}=1)= \frac{1}{4} + \frac{(5)^{1/2}}{4}$
But i'm feel it's wrong.
It is not
The expectation is correct.
$$\def\Chi{\mathcal{X}}\mathsf E(\Chi)~{=1m+(-1)(1-m)\\=2m-1}$$
The variance is correct:
$$\mathcal {Var}(\Chi)~{ =1^2m+(-1)^2(1-m)-(2m-1)^2\\=1-4m^2+4m-1\\=4m-4m^2}$$
So indeed you need to solve $0=4m^2-2m-1$ for $m\in[0..1]$.$$m~{=\dfrac{2\pm\surd(4+16)}{8}\\=\dfrac{1\pm\surd 5}{4}}$$
Of which only one root lies within the required interval:$$m=\dfrac{1+\surd 5}{4}$$