91Ó°ÊÓ

Referring to the sec. 1.10, there are n letters and n envelopes. The letters have to be put into the correct envelopes.

Let’s consider some events, such that,

\({M_n} = {X_1} + {X_2} + \cdots + {X_n}\)= the number of correct matches

Let’s assume that\({X_i}\)follows Bernoulli distribution. So, the variable is either 0 if placed wrong envelop or 1 if placed right envelop.

So, for\(\left( {{X_1} = 1} \right)\)the probability mass function is\(\Pr \left( {{X_1} = 1} \right) = \frac{1}{n}\)

And for\(\left( {{X_1} = 0} \right)\)the p.m.f is\(\Pr \left( {{X_1} = 0} \right) = 1 - \frac{1}{n} = \left( {\frac{{n - 1}}{n}} \right)\).

The expected value of\({X_1}\)is,

\(\begin{aligned}{}E\left( {{X_1}} \right) = 1 \times \frac{1}{n} + 0 \times \left( {\frac{{n - 1}}{n}} \right)\\ = \frac{1}{n}\end{aligned}\)

Similarly, there can be obtained the expected value of \({X_2},{X_3}\) and so on.

The expected number of correct matches is,

\(\begin{aligned}{}E\left( {{M_n}} \right) = \sum\limits_{k = 0}^n {k \times \Pr \left( {X = k} \right)} \\ = E\left( {{X_1} + {X_2} + \cdots + {X_3}} \right)\\ = \sum\limits_{i = 1}^n {E\left( {{X_i}} \right)} \\ = n \times \frac{1}{n}\\ = 1\end{aligned}\)

Now,

\(\begin{aligned}{}E\left( {M_n^2} \right) = \sum\limits_{k = 0}^n {{k^2}P\left( {{M_n} = k} \right)} \\ = \sum\limits_{k = 1}^{n - 2} {{k^2}\left( {\frac{1}{{k!}}\sum\limits_{i = 2}^{n - k} {\frac{{{{\left( { - 1} \right)}^i}}}{{i!}}} } \right) + {n^2}\frac{1}{{n!}}} \\ = \sum\limits_{k = 1}^{n - 2} {\left( {\frac{k}{{\left( {k + 1} \right)!}}\sum\limits_{i = 2}^{n - k} {\frac{{{{\left( { - 1} \right)}^i}}}{{i!}}} } \right) + \frac{n}{{\left( {n - 1} \right)!}}} \\ = \sum\limits_{k = 0}^{n - 3} {\left( {\frac{{k + 1}}{{k!}}\sum\limits_{i = 2}^{n - 1 - k} {\frac{{{{\left( { - 1} \right)}^i}}}{{i!}}} } \right) + \frac{n}{{\left( {n - 1} \right)!}}} \\ = \sum\limits_{k = 0}^{n - 3} {k\left( {\frac{1}{{k!}}\sum\limits_{i = 2}^{n - 1 - k} {\frac{{{{\left( { - 1} \right)}^i}}}{{i!}}} } \right) + \sum\limits_{k = 0}^{n = 3} {\left( {\frac{1}{{k!}}\sum\limits_{i = 2}^{n - 1 - k} {\frac{{{{\left( { - 1} \right)}^i}}}{{i!}}} } \right)} } + \frac{{n - 1 + 1}}{{\left( {n - 1} \right)!}}\\ = \left( {\sum\limits_{k = 0}^{n - 3} {k\left( {\frac{1}{{k!}}\sum\limits_{i = 2}^{n - 1 - k} {\frac{{{{\left( { - 1} \right)}^i}}}{{i!}}} } \right)} + \frac{{n - 1}}{{\left( {n - 1} \right)!}}} \right) + \left( {\sum\limits_{k = 0}^{n - 3} {k\left( {\frac{1}{{k!}}\sum\limits_{i = 2}^{n - 1 - k} {\frac{{{{\left( { - 1} \right)}^i}}}{{i!}}} } \right)} + \frac{1}{{\left( {n - 1} \right)!}}} \right)\\ = E\left( {{M_{n - 1}}} \right) + \sum\limits_{k = 0}^{n - 1} {\Pr \left( {{M_{n - 1}} = k} \right)} \\ = 1 + 1\\ &= 2\end{aligned}\)

So, the variance of\({M_n}\)is,

\(\begin{aligned}{}Var\left( {{M_n}} \right) = E\left( {M_n^2} \right) - {\left( {E\left( {{M_n}} \right)} \right)^2}\\ = 2 - 1\\ = 1\end{aligned}\)

Therefore, the variance of the number of letters that are placed in the correct envelopes is 1.