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Prove that if a sequence\({Z_1},{Z_2},...\)converges to a constant b in quadratic mean, then the sequence also converges to b in probability.

Short Answer

Expert verified

It is proved that the given sequence converges to b in probability.

Step by step solution

01

Given information

The sequence \({Z_1},{Z_2},...\)converges to b in the quadratic mean.

02

Proving that the sequence converges to b in probability

Since,

\(\left| {{Z_n} - b} \right| \le \left| {{Z_n} - E\left( {{Z_n}} \right)} \right| + \left| {E\left( {{Z_n}} \right) - b} \right|,\)

Then for any value of\( \in > 0\),

\(\begin{array}{c}P\left( {\left| {{Z_n} - b} \right| < \in } \right) \ge \Pr \left( {\left| {{Z_n} - E\left( {{Z_n}} \right)} \right|} \right) + \left( {\left| {E\left( {{Z_n}} \right) - b} \right| < \in } \right)\\ = \Pr \left( {\left| {{Z_n} - E\left( {{Z_n}} \right)} \right|} \right) < \in - \left| {E\left( {{Z_n}} \right) - b} \right|\end{array}\)

Since,

\(\mathop {\lim }\limits_{n \to \infty } E\left( {{Z_n}} \right) = b\)

Therefore, for sufficiently large values of n, it will be true that

\( \in - \left| {E\left( {{Z_n}} \right) - b} \right| > 0\)

Hence,

By the Chebyshev inequality, the final probability will be at least as large as

\(1 - \frac{{Var\left( {{Z_n}} \right)}}{{{{\left[ { \in - \left| {E\left( {{Z_n}} \right) - b} \right|} \right]}^2}}}\)

Again,

\(\mathop {\lim }\limits_{n \to \infty } V\left( {{Z_n}} \right) = 0\)and \(\mathop {\lim }\limits_{n \to \infty } {\left[ { \in - \left| {E\left( {{Z_n}} \right) - b} \right|} \right]^2} = { \in ^2}\).

Therefore,

\(\begin{array}{c}\mathop {\lim }\limits_{n \to \infty } \Pr \left[ {\left| {E\left( {{Z_n}} \right) - b} \right| < \in } \right] \ge \mathop {\lim }\limits_{n \to \infty } \left\{ {1 - \frac{{Var\left( {{Z_n}} \right)}}{{{{\left[ { \in - \left| {E\left( {{Z_n}} \right) - b} \right|} \right]}^2}}}} \right\}\\ = 1\end{array}\)

Which means that \({Z_n}\xrightarrow{P}b\) .

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