91Ó°ÊÓ

A sequence of random variables\({Z_1},{Z_2},...\)converges to b in quadratic mean if \(\mathop {\lim }\limits_{n \to \infty } E\left[ {{{\left( {{Z_n} - b} \right)}^2}} \right] = 0\).

\(\begin{array}{c}E\left[ {{{\left( {{Z_n} - b} \right)}^2}} \right] = E\left[ {Z_{_n}^2 - 2{Z_n}b + {b^2}} \right]\\ = E\left[ {Z_n^2} \right] - 2bE\left( {{Z_n}} \right) + {b^2}\end{array}\)

Since,

\(V\left[ {{Z_n}} \right] = E\left( {Z_n^2} \right) - {\left[ {E\left( {{Z_n}} \right)} \right]^2}\)

So,

\(E\left( {Z_n^2} \right) = V\left[ {{Z_n}} \right] + {\left[ {E\left( {{Z_n}} \right)} \right]^2}\)

Substitute in above expression

\(\begin{array}{c}E\left[ {{{\left( {{Z_n} - b} \right)}^2}} \right] = V\left( {{Z_n}} \right) + {\left[ {E\left( {{Z_n}} \right)} \right]^2} - 2bE\left( {{Z_n}} \right) + {b^2}\\ = V\left( {{Z_n}} \right) + {\left[ {E\left( {{z_n}} \right) - b} \right]^2}\end{array}\)

Thus, the limit \(n \to \infty \)of\(E\left[ {{{\left( {{Z_n} - b} \right)}^2}} \right]\) will be 0 if and only if the limit of each of the two terms on the right side will be 0.

That is:

\(\begin{array}{c}{\mathop {\lim \left[ {E\left( {{Z_n}} \right) - b} \right]}\limits_{n \to \infty } ^2} = 0\\\mathop {\lim }\limits_{x \to \infty } E\left( {{Z_n}} \right) = b\end{array}\)

And,

\(\mathop {\lim }\limits_{n \to \infty } V\left( {{Z_n}} \right) = 0\)

Thus, the required result is proved.