Question

It$X$has, a mean$\mu$and standard deviation$\sigma$, the ratio$r=\left|\mu \right|/\sigma$is called the measurement signal-to-noise ratio$X$. The idea is that $X$can be expressed as$X=\mu +(X-\mu )$, $\mu$representing the signal and $X-\mu$the noise. If we define$\left|\right(X-\mu )/\mu |=D$it as the relative deviation $X$from its signal (or mean)$\mu$, show that for$\alpha >0$,

$P\{D\le \alpha \}\ge 1-\frac{1}{r^2{\alpha }^2}$.

P{D≤α}≥1−1r2α2

Short answer

Since $\left|\mu \right|\alpha >0$using Chebyshev's inequality we get:

$P\left\{\right|X-\mu |>|\mu \left|\alpha \right\}\le \frac{{\sigma }^2}{{\mu }^2{\alpha }^2}.$

Therefore,

$P\{D\le \alpha \}=P\left\{\right|X-\mu |\le |\mu \left|\alpha \right\}\ge 1-\frac{{\sigma }^2}{{\mu }^2{\alpha }^2}\stackrel{r=\frac{\left|u\right|}{=}}{=}1-\frac{1}{r^2{\alpha }^2}$.

Step-by-step solution

Given Information.

$X$has to mean$\mu$and standard deviation$\sigma$, the ratiolocalid="1649850029051" $r=\left|\mu \right|/\sigma$called the measurement signal-to-noise ratio of X.

Explanation.

Assume that the random variable $X$has mean $\mu$and standard deviation$\sigma$, and let$\alpha >0$. Then,

$P\{D\le \alpha \}=P\left\{\left|\frac{\mid X-\mu }{\mu }\right|\le \alpha \right\}=P\left\{\right|X-\mu |\le |\mu \left|\alpha \right\}$

Since localid="1649850016019" $\left|\mu \right|\alpha >0$using Chebyshev's inequality we get:

$P\left\{\right|X-\mu |>|\mu \left|\alpha \right\}\le \frac{{\sigma }^2}{{\mu }^2{\alpha }^2}$.

Since

$P\left\{\right|X-\mu |>|\mu \left|\alpha \right\}=1-P\left\{\right|X-\mu |\le |\mu \left|\alpha \right\}$,

therefore

$$\begin{gathered} P\{D\le \alpha \}=P\left\{\right|X-\mu |\le |\mu \left|\alpha \right\}\ge 1-\frac{{\sigma }^2}{{\mu }^2{\alpha }^2} \\ \Downarrow r=\frac{\left|\mu \right|}{\sigma } \\ P\{D\le \alpha \}\ge 1-\frac{1}{r^2{\alpha }^2} \end{gathered}$$.