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Thursday, 7 August 2014

chapter-21 Worked Out Examples

    Example: 19   

Show that {\rm{Var}}(X) = \left\langle {{X^2}} \right\rangle  - {\left\langle X \right\rangle ^2}
Solution: 19

{\rm{Var}}(X) = \left\langle {{{\left( {X - \left\langle X \right\rangle } \right)}^2}} \right\rangle \;
 = \sum\limits_{i = 1}^n {{{\left( {{x_i} - \left\langle X \right\rangle } \right)}^2}{p_i}}  {where the symbols have their usual meaning}
 = \sum\limits_{i = 1}^n {\left( {x_i^2 + {{\left\langle X \right\rangle }^2} - 2{x_i}\left\langle X \right\rangle } \right){p_i}}
 = \sum\limits_{i = 1}^n {{p_i}x_i^2}  + \sum\limits_{i = 1}^n {{p_i}{{\left\langle X \right\rangle }^2}}  - 2\;\sum\limits_{i = 1}^n {{x_i}{p_i}\left\langle X \right\rangle }
Since \left\langle X \right\rangle \,{\rm{and}}\,{\left\langle X \right\rangle ^2} are constants, they can be taken out of the summation in the second and third terms. Also, note that
\sum\limits_{i = 1}^n {{p_i}x_i^2}  = \left\langle {{X^2}} \right\rangle ,\;\;\sum\limits_{i = 1}^n {{p_i} = 1} ,\;\;\sum\limits_{i = 1}^n {{p_i}{x_i}}  = \left\langle X \right\rangle \;
so that,
{\rm{Var}}(X) = \;\left\langle {{X^2}} \right\rangle  + {\left\langle X \right\rangle ^2} - 2{\left\langle X \right\rangle ^2}
 = \left\langle {{X^2}} \right\rangle  - {\left\langle X \right\rangle ^2}
which proves the assertion.
The relation is important and useful since it gives us the variance directly in terms of \left\langle X \right\rangle  and \left\langle {{X^2}} \right\rangle . You are urged to try using this relation to calculate variance in the examples of variance we’ve discussed earlier.
     Example: 20  
Two cards are drawn simultaneously from a well-shuffled deck of 52 cards. Find the mean and variance of the number of kings.
Solution: 20

The number of kings is the random variable here. Call it X. The values that X can take are 012. The probabilities of the various values are easily calculated:
P(X = 0) = \dfrac{{^{48}{C_2}}}{{^{52}{C_2}}} = \dfrac{{188}}{{221}}
P(X = 1) = \dfrac{{^4{C_1} \times {\,^{48}}{C_1}}}{{^{52}{C_2}}} = \dfrac{{32}}{{221}}
P(X = 2) = \dfrac{{^4{C_2}}}{{^{52}{C_2}}} = \dfrac{1}{{221}}
The PD of X is therefore
The mean is simply
\left\langle X \right\rangle \; = 0 \times \dfrac{{188}}{{221}} + 1 \times \dfrac{{32}}{{221}} + 2 \times \dfrac{1}{{221}}
 = \dfrac{{34}}{{221}}
To calculate the variance, we first calculate \left\langle {{X^2}} \right\rangle :
\left\langle {{X^2}} \right\rangle \; = \sum\limits_{i = 1}^n {x_i^2} {p_i}
 = {0^2} \times \dfrac{{188}}{{221}} + {1^2} \times \dfrac{{32}}{{221}} + {2^2} \times \dfrac{1}{{221}} = \dfrac{{36}}{{221}}
Thus, the variance is
{\rm{Var}}(X) = \;\left\langle {{X^2}} \right\rangle  - {\left\langle X \right\rangle ^2}
 = \dfrac{{36}}{{221}} - {\left( {\dfrac{{34}}{{221}}} \right)^2}
 = \dfrac{{6800}}{{48841}}
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