Question

It $X$has a variance${\sigma }^2$, then σ, the positive square root of the variance, is called the standard deviation. It $X$has to mean $\mu$and standard deviation$\sigma$, to show that$P\left\{\right|X-\mu |\ge k\sigma \}\le \frac{1}{k^2}.$

Short answer

Since$|X-\mu |\ge 0$,

$$\begin{gathered} {\left(\frac{|X-\mu |}{k\sigma }\right)}^2\ge \frac{|X-\mu |}{k\sigma }\ge I_{\left\{\right|X-\mu |\ge k\sigma \}} \\ \Downarrow \\ \frac{1}{k^2}=E^{\delta }\left[{\left(\frac{|X-\mu |}{k\sigma }\right)}^2\right]\ge E\left[I_{\left\{\right|X-\mu |>k\sigma \}}\right]=P\left\{\right|X-\mu |\ge k\sigma \} \end{gathered}$$

On the other hand, this result can be proven directly using Chebyshev's inequality. Since$k\sigma >0$,

$P\left\{\right|X-\mu |\ge k\sigma \}\le \frac{{\sigma }^2}{(k\sigma {)}^2}=\frac{1}{k^2}.$

Step-by-step solution

Given Information.

It $X$has a variance${\sigma }^2$, then$\sigma$, the positive square root of the variance is called the standard deviation. It $X$has to mean $\mu$and standard deviation$\sigma$,

Explanation.

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

Let

$$\begin{gathered} I_{\left\{\right|X-\mu |\ge k\sigma \}}=1\text{if}|X-\mu |\ge k\sigma \\ 0,\text{otherwise} \end{gathered}$$

Then

$P\left\{\right|X-\mu |\ge k\sigma \}=E\left[I_{\left\{\right|X-\mu |\ge k\sigma \}}\right]$

Since$|X-\mu |>0$note that.

$$\begin{gathered} {\left(\frac{|X-\mu |}{k\sigma }\right)}^2\ge \frac{|X-\mu |}{k\sigma }\ge I_{\left\{\right|X-\mu |>k\sigma \}} \\ E\left[{\left(\frac{|X-\mu |}{k\sigma }\right)}^2\right]\ge E\left[I_{\left\{\right|X-\mu |\ge k\sigma \}}\right] \\ \frac{1}{k^2{\sigma }^2}{\overbrace{E\left[|X-\mu {|}^2\right]}}^{={\sigma }^2}\ge P\left\{\right|X-\mu |\ge k\sigma \} \\ \mathrm{\mathbb{N}} \\ \frac{1}{k^2}\ge P\left\{\right|X-\mu |\ge k\sigma \} \end{gathered}$$

Explanation.

On the other hand, this result can be proven directly using Chebyshev's inequality. Since$k\sigma >0$,

$P\left\{\right|X-\mu |\ge k\sigma \}\le \frac{{\sigma }^2}{(k\sigma {)}^2}=\frac{1}{k^2}.$