State and prove chebyshev's inequality theorem in probability
Answers
Answer:
1. In the case of a discrete random variable
\displaystyle X
, the probability density function is
\displaystyle f(x)
,
\displaystyle \sigma^2 = \sum\limits_{i=1} (x_i-\mu)^2f(x_i) \hspace{8mm} \textcircled 1
For those
\displaystyle x_j
in the domain of
\displaystyle \mid X-\mu \mid \geqslant k\sigma
,
\displaystyle (x_j-\mu)^2 \geqslant k^2\sigma^2
. So,
\displaystyle \textcircled 1 \geqslant \sum\limits_{j, \mid X-\mu \mid \geqslant k\sigma} (x_j-\mu)^2f(x_j) \\\\ \geqslant \sum\limits_{j, \mid X-\mu \mid \geqslant k\sigma} k^2\sigma^2f(x_j) \\\\ = k^2\sigma^2\sum\limits_{j, \mid X-\mu \mid \geqslant k\sigma} f(x_j) \\\\ = k^2\sigma^2 P \left(\mid X-\mu \mid \geqslant k\sigma \right)
Then we can get the inequality
\displaystyle P\left(\mid X-\mu \mid \geqslant k\sigma \right) \leqslant \frac{1}{k^2}.
2. In the case of a continuous random variable
\displaystyle X
,
\displaystyle \sigma^2 = \int_{-\infty}^{\infty} (x-\mu)^2f(x)dx \\\\ = \int_{\mid X-\mu \mid \geqslant k\sigma} (x-\mu)^2f(x)dx + \int_{\mid X-\mu \mid < k\sigma} (x-\mu)^2f(x)dx \\\\ \geqslant \int_{\mid X-\mu \mid \geqslant k\sigma} (x-\mu)^2f(x)dx \hspace{8mm} \textcircled 2
Just like discrete distribution discussed, for those
\displaystyle x
in the domain of
\displaystyle \mid X-\mu \mid \geqslant k\sigma
,
\displaystyle (x-\mu)^2 \geqslant k^2\sigma^2
. So,
\displaystyle \textcircled 2 \geqslant \int_{\mid X-\mu \mid \geqslant k\sigma} k^2\sigma^2f(x)dx \\\\ = k^2\sigma^2 \int_{\mid X-\mu \mid \geqslant k\sigma} f(x)dx \\\\ = k^2\sigma^2 P \left(\mid X-\mu \mid \geqslant k\sigma \right)
And we can get the same inequality
\displaystyle P\left(\mid X-\mu \mid \geqslant k\sigma \right) \leqslant \frac{1}{k^2}.