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*Continuous case:<br> | *Continuous case:<br> | ||
| + | |||
| + | Let X be a random variable with mean <math>\mu</math> and variance <math>\sigma^2</math> | ||
Consider two functions <span class="texhtml">''g''<sub>''1''</sub>(''x'')</span> and <span class="texhtml">''g''<sub>''2''</sub>(''x'')</span> | Consider two functions <span class="texhtml">''g''<sub>''1''</sub>(''x'')</span> and <span class="texhtml">''g''<sub>''2''</sub>(''x'')</span> | ||
| − | |||
| − | |||
| − | |||
| − | |||
<math>g_{1}(x)= 1_{\{x: \mid x-\mu\mid \geq \varepsilon\}}(x)</math> | <math>g_{1}(x)= 1_{\{x: \mid x-\mu\mid \geq \varepsilon\}}(x)</math> | ||
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<math>g_{2}(x)=\frac{(x-\mu)^2}{\varepsilon^2}</math> | <math>g_{2}(x)=\frac{(x-\mu)^2}{\varepsilon^2}</math> | ||
| + | Clearly, | ||
| + | |||
| + | <math>g_{2}(x)-g_{1}(x) \geq 0</math> | ||
| + | |||
| + | <math>E[g_{2}(x)-g_{1}(x) ] \geq 0</math> | ||
| + | |||
| + | Consider, | ||
| + | |||
| + | <math>E[g_{2}(x)-g_{1}(x) ] \geq 0 = E[g_{2}(x)]-E[g_{1}(x)]</math> | ||
| + | |||
| + | where, | ||
| + | <math>E[g_{2}(x)]=E[\frac{(x-\mu)^2}{\varepsilon^2}] = \frac{1}{\varepsilon^2} var(X) = \frac{]sigma^2}{\varepsilon^2}</math> | ||
| + | |||
| + | and | ||
| + | |||
| + | <math>E[g_{1}(x)] = P \{\mid x - \mu\mid \geq \varepsilon \}</math> | ||
| + | |||
| + | Thus we get, | ||
| + | |||
| + | <math>\frac{\sigma^2}{\varepsilon^2} - P\{\mid X - \mu \mid \geq \varepsilon\} \geq 0</math> | ||
| + | |||
| + | Therefore, | ||
| + | |||
| + | <math>P\{\mid X - \mu \mid \geq \varepsilon\} \leq \frac{\sigma^2}{\varepsilon^2}</math> | ||
Revision as of 12:30, 26 January 2014
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2012
Problem 2
Problem statement: Let X be a continuous or discrete random variable with mean μ and variance σ2. Then, $ \forall \varepsilon >0 $, we have
$ P(|X-\mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2} $
$ \color{blue}\text{Solution 1:} $
- Continuous case:
Let X be a random variable with mean $ \mu $ and variance $ \sigma^2 $
Consider two functions g1(x) and g2(x)
$ g_{1}(x)= 1_{\{x: \mid x-\mu\mid \geq \varepsilon\}}(x) $
$ g_{2}(x)=\frac{(x-\mu)^2}{\varepsilon^2} $
Clearly,
$ g_{2}(x)-g_{1}(x) \geq 0 $
$ E[g_{2}(x)-g_{1}(x) ] \geq 0 $
Consider,
$ E[g_{2}(x)-g_{1}(x) ] \geq 0 = E[g_{2}(x)]-E[g_{1}(x)] $
where, $ E[g_{2}(x)]=E[\frac{(x-\mu)^2}{\varepsilon^2}] = \frac{1}{\varepsilon^2} var(X) = \frac{]sigma^2}{\varepsilon^2} $
and
$ E[g_{1}(x)] = P \{\mid x - \mu\mid \geq \varepsilon \} $
Thus we get,
$ \frac{\sigma^2}{\varepsilon^2} - P\{\mid X - \mu \mid \geq \varepsilon\} \geq 0 $
Therefore,
$ P\{\mid X - \mu \mid \geq \varepsilon\} \leq \frac{\sigma^2}{\varepsilon^2} $
$ \color{blue}\text{Solution 2:} $
- Discrete Case:
Let pX(x) be the pmf of X. The probability that X differs from μ by at least $ \varepsilon $ is
$ P(|X-\mu| \geq \varepsilon)= \sum_{|X-\mu| \geq \varepsilon}p_{X}(x) $
Based on the definition of the variance, we have
| σ2 = | ∑ | (x − μ)2pX(x) |
| x |
Let a set $ A= \{ x|\,|x-\mu| \geq \varepsilon \} $. We have
$ \sigma^2 = \sum_{x}(x-\mu)^2 p_{X}(x)= \sum_{x \in A}(x-\mu)^2 p_{X}(x)+\sum_{x \notin A}(x-\mu)^2 p_{X}(x) $
$ \Rightarrow\sigma^2 \geq \sum_{x \in A}(x-\mu)^2 p_{X}(x) $
Since, in set A, we have $ |x-\mu| \geq \varepsilon $, we have
$ \Rightarrow\sigma^2 \geq \sum_{x \in A}\varepsilon^2 p_{X}(x)= \varepsilon^2 \sum_{x \in A}p_{X}(x)=\varepsilon^2 P(x \in A) =\varepsilon^2 P(|X-\mu| \geq \varepsilon) $
That is
$ P(|X-\mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2} $
- Continuous Case:
Let fX(x) be the pdf of X.
$ \sigma^2=\int_{-\infty}^{\infty}(x-\mu)^2f_{X}(x) \,dx \geq \int_{-\infty}^{\mu-\varepsilon}(x-\mu)^2f_{X}(x) \,dx+ \int_{\mu+\varepsilon}^{\infty}(x-\mu)^2f_{X}(x) \,dx $
The last inequality holds since we integrate a positive function. Since $ x \leq \mu-\varepsilon $ or $ x \geq \mu+\varepsilon $
$ \Rightarrow |x-\mu| \geq \varepsilon \Rightarrow (x-\mu)^2 \geq \varepsilon^2 $
Based on the above equation, we have
$ \sigma^2 \geq \int_{-\infty}^{\mu-\varepsilon}\varepsilon^2 f_{X}(x) \,dx+ \int_{\mu+\varepsilon}^{\infty} \varepsilon^2 f_{X}(x) \,dx $
$ = \varepsilon^2 \left( \int_{-\infty}^{\mu-\varepsilon}f_{X}(x) \,dx+ \int_{\mu+\varepsilon}^{\infty} f_{X}(x) \,dx \right) = \varepsilon^2 P \bigg( X \leq (\mu-\varepsilon)\, \text{or} \, X \geq (\mu+\varepsilon) \bigg) = \varepsilon^2 P(|X-\mu| \geq \varepsilon) $
$ \Rightarrow P(|X-\mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2} $
