If we have two gaussian random variables $x$ and $y$ where:
- $x$ has mean $0$ and $a \leq$ variance $\leq b$ and
- $y$ has mean $0$ and variance $1$
is it reasonable to conclude the following statement?
$P(c\cdot x <1) \leq P(\sqrt{a} \cdot y < \frac{1}{c})$
where $c$ is just some real number.
I think this first conclusion would seem reasonable based on what I was reading here, but it seemed conditional so I just wanted to confirm it and if possible gain more insight.
By the same logic, if we have the gaussian random variables
- $x_i$ have mean $0$ and $a \leq$ variance $\leq b$ and
- $y_i$ has mean $0$ and variance $1$
and suppose $u_i$ are an orthonormal basis. Could we conclude the following statement as well?
$P(c \cdot || \sum_{i} x_i \cdot u_i || > d) \leq P(c \cdot \sqrt{b} \sqrt{ \sum_{i} y_i ^2} > d)$
where $c$ and $d$ are just some real numbers.
I think this would make sense from the above combined with what I read here, but again I would love to confirm this and I'm grateful for any insight.
I would make another approach. The random variables are distributed as follows:
$X\sim \mathcal N(0,\sigma_x^2), a\leq \sigma_x^2\leq b, \quad Y\sim \mathcal N(0,1)$. The inequality for the probabilities can be written as
$P\left(X<\frac1c \right)\leq P\left(Y<\frac{1}{c\cdot \sqrt a} \right)$.
I assume that $c\neq 0$. With the standardization of the random variables $Z=\frac{X-\mu}{\sigma}$ and the cdf of the standard normal distribution $\Phi(z)$ we obtain
$$\Phi\left(\frac{\frac1c-0}{\sigma_x} \right)\leq \Phi\left(\frac{\frac1{c\sqrt a}-0}{1} \right)$$
$$\Phi\left(\frac{1}{c\cdot \sigma_x} \right)\leq \Phi\left(\frac1{c\sqrt a} \right)$$
Taking the inverse funtion, $\Phi^{-1}(\cdot )$. The inequality still holds, since $\Phi(z)$ is a monotone increasing function.
$\frac{1}{c\cdot \sigma_x} \leq \frac1{c\sqrt a} $
$$\frac{1}{\sigma_x} \leq \frac1{\sqrt a}\Rightarrow \sqrt a\leq \sigma_x\Rightarrow a\leq \sigma_x^2$$ This condition is fulfilled since it is one of the defined condition in the exercise. Therefore the statement at the beginning is true.