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Suppose that X is a random variable for which \({\bf{E}}\left( {\bf{X}} \right){\bf{ = 10}}\) ,\({\bf{P}}\left( {{\bf{X}} \le {\bf{7}}} \right){\bf{ = 0}}{\bf{.2}}\,\,{\bf{and}}\,{\bf{P}}\left( {{\bf{X}} \ge {\bf{13}}} \right){\bf{ = 0}}{\bf{.3}}\) .Prove that,\({\bf{Var}}\left( {\bf{X}} \right) \ge \frac{{\bf{9}}}{{\bf{2}}}\) .

Short Answer

Expert verified

\({\rm{Var}}\left( X \right) \ge \frac{9}{2}\) .

Step by step solution

01

Given information

For a random variable X ,\(P\left( {X \le 7} \right) = 0.2\,\,{\rm{and}}\,P\left( {X \ge 13} \right) = 0.3\)and also \(E\left( X \right) = 10\).We need to proof the inequality \({\rm{Var}}\left( X \right) \ge \frac{9}{2}\)

02

Proof of the inequality

Chebyshev inequality

Let X be the random variable for which \({\rm{Var}}\left( X \right)\)exists. Then every for every number \(t > 0\) we have the following inequality

\(P\left( {\left| {X - E\left( X \right)} \right| \ge t} \right) \le \frac{{{\rm{Var}}\left( X \right)}}{{{t^2}}}\)

Now,

\(P\left( {\left| {X - E\left( X \right)} \right| \ge t} \right) = P\left( { - t \le \left( {X - E\left( X \right)} \right) \ge t} \right)\)

Now

\(P\left( {X \ge 13} \right) = 0.3\)implies

\(\begin{aligned}{}t + E\left( X \right) &= 13\\t + 10 &= 13\\t &= 3\end{aligned}\)

Putting this value of t, we get

\(\begin{aligned}{}\frac{{{\rm{Var}}\left( X \right)}}{9} \ge P\left( {\left| {X - 10} \right| \ge 3} \right)\\ &= P\left( {X \le 7\,{\rm{or}}\,X \ge 10} \right)\\ &= P\left( {X \le 7} \right) + P\left( {X \ge 10} \right)\\ &= 0.2 + 0.3\\ &= 0.5\\ &= \frac{1}{2}\end{aligned}\)

Thus, \({\rm{Var}}\left( X \right) \ge \frac{9}{2}\) .

Hence the proof.

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