Problem source

Let $X$ be a random variable which takes only the finite number of different possible real values $x_{1},x_{2},\ldots,x_{n}.$ Define the expectation $\mathbb{E}(X)$ and the variance $\var(X)$
of $X$. Show that , if $a$ and $b$ are real numbers, then $\E(aX+b)=a\E(X)+b$
and express $\var(aX+b)$ similarly in terms of $\var(X)$. 
Let $\lambda$ be a positive real number. By considering the contribution to $\var(X)$ of those $x_{i}$ for which $\left|x_{i}-\E(X)\right|\geqslant\lambda,$ or otherwise, show that 
\[
\mathrm{P}\left(\left|X-\E(X)\right|\geqslant\lambda\right)\leqslant\frac{\var(X)}{\lambda^{2}}\,.
\]
Let $k$ be a real number satisfying $k\geqslant\lambda.$ If $\left|x_{i}-\E(X)\right|\leqslant k$ for all $i$, show that 
\[
\mathrm{P}\left(\left|X-\E(X)\right|\geqslant\lambda\right)\geqslant\frac{\var(X)-\lambda^{2}}{k^{2}-\lambda^{2}}\,.
\]

Solution source

Definition: $\displaystyle \mathbb{E}(X) = \sum_{i=1}^n x_i \mathbb{P}(X = x_i)$

Definition: $\displaystyle \mathrm{Var}(X) = \sum_{i=1}^n (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i)$

Claim: $\mathbb{E}(aX+b) = a\mathbb{E}(X)+b$
Proof: 
\begin{align*}
\mathbb{E}(aX+b) &=  \sum_{i=1}^n (ax_i+b) \mathbb{P}(X = x_i) \\
&= a\sum_{i=1}^n x_i \mathbb{P}(X = x_i) + b\sum_{i=1}^n  \mathbb{P}(X = x_i)\\
&= a \mathbb{E}(X) + b
\end{align*}

Claim: $\mathrm{Var}(aX+b) = a^2 \mathrm{Var}(X)$

Claim: $\mathrm{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right)\leqslant\frac{\mathrm{var}(X)}{\lambda^{2}}$

Proof: 
\begin{align*}
 \mathrm{Var}(X) &= \sum_{i=1}^n (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\
&\geq \sum_{|x_i - \mathbb{E}(X)| \geq  \lambda} (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\ 
&\geq \sum_{|x_i - \mathbb{E}(X)| \geq  \lambda} \lambda^2 \mathbb{P}(X = x_i) \\ 
&= \lambda^2 \sum_{|x_i - \mathbb{E}(X)| \geq  \lambda}  \mathbb{P}(X = x_i) \\ 
&= \lambda^2 \mathrm{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right)
\end{align*}

Claim: \[
\mathrm{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right)\geqslant\frac{\mathrm{var}(X)-\lambda^{2}}{k^{2}-\lambda^{2}}\,.
\]

Proof:

\begin{align*}
&& \mathrm{Var}(X) &= \sum_{i=1}^n (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\
&&&= \sum_{|x_i - \mathbb{E}(X)| \geq  \lambda} (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) + \sum_{|x_i - \mathbb{E}(X)| < \lambda} (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\
&&& \leq  \sum_{|x_i - \mathbb{E}(X)| \geq  \lambda} k^2 \mathbb{P}(X = x_i) + \sum_{|x_i - \mathbb{E}(X)| < \lambda} \lambda^2 \mathbb{P}(X = x_i) \\
&&&= k^2 \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) + \lambda^2 \mathbb{P}\left(\left|X-\mathrm{E}(X)\right| < \lambda\right) \\
&&&=  k^2 \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) + \lambda^2(1- \mathbb{P}\left(\left|X-\mathrm{E}(X)\right| \leq \lambda\right) \\
&&&= (k^2 - \lambda^2) \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) + \lambda^2 \\
\Rightarrow&& \frac{\mathrm{Var}(X)-\lambda^2}{k^2 - \lambda^2} &\leq \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) 
\end{align*}


[Note: This result is known as Chebyshev's inequality, and is an important starting point to understanding the behaviour of tails of random variables]