problem 12

6.1.3 problem 12

(a). Note that, when $Y$ is a discrete r.v. whose possible values are $y_{1}, . . ., y_{n}$ ,

\begin{matrix} P ({(\frac{1}{n} n \sum j = 1 y_{j})}_{j}^{2} \leq \frac{1}{n} n \sum j = 1 y_{j}^{2}) & = P ({(\frac{1}{n} n \sum j = 1 y_{j})}_{j}^{2} - \frac{1}{n} n \sum j = 1 y_{j}^{2} \leq 0) = P (\frac{1}{n} n \sum j = 1 y_{j}^{2} - {(\frac{1}{n} n \sum j = 1 y_{j})}_{j}^{2} \geq 0) = P (E (Y^{2}) - E {(Y)}^{2} \geq 0) = P (Var (Y) \geq 0) = 1, \end{matrix}

where $Var (Y) = 0$ if and only if $y_{1} = . . . = y_{n}$ . Since $X_{j}$ are continuous r.v.s, $P (X_{1} = . . . = X_{n}) = P (T_{1} = T_{2}) = 0$ , therefore,

\begin{matrix} P (T_{1} \leq T_{2}) & = P (T_{1} < T_{2}) = P (T_{2} - T_{1} > 0) = P (\frac{1}{n} n \sum j = 1 X_{j}^{2} - {(\frac{1}{n} n \sum j = 1 X_{j})}_{j}^{2} > 0) = 1 . \end{matrix}

(b). Note that we want to estimate $sigm a^{2}$ , hence $θ = c^{2}$ .
To find the bias of $T_{1}$ , we want to find $E (T_{1}) - c^{2}$ .

\begin{matrix} E (T_{1}) & = E ({¯ ¯ ¯ X}_{n}^{2}) = Var ({¯ ¯ ¯ X}_{n}) + {(E ({¯ ¯ ¯ X}_{n}))}_{n}^{2} = Var (\frac{1}{n} n \sum j = 1 X_{j}) + {(E (\frac{1}{n} n \sum j = 1 X_{j}))}_{j}^{2} = \frac{1}{n^{2}} Var (n \sum j = 1 X_{j}) + {(\frac{1}{n} E (n \sum j = 1 X_{j}))}_{j}^{2} = \frac{1}{n^{2}} Var (n \sum j = 1 X_{j}) + \frac{1}{n^{2}} {(E (n \sum j = 1 X_{j}))}_{j}^{2} \end{matrix}

Since each $X_{j}$ is mutually independent, and by linearity of expectation,

\begin{matrix} E (T_{1}) & = \frac{1}{n^{2}} n \sum j = 1 Var (X_{j}) + \frac{1}{n^{2}} {(n \sum j = 1 E (X_{j}))}_{j}^{2} = \frac{1}{n^{2}} n \sum j = 1 σ^{2} + \frac{1}{n^{2}} {(n \sum j = 1 c)}^{2} = \frac{1}{n^{2}} \cdot n σ^{2} + \frac{1}{n^{2}} \cdot {(nc)}^{2} = \frac{σ^{2}}{n} + c^{2} \end{matrix}

Hence the bias of $T_{1}$ is

\frac{σ^{2}}{n} + c^{2} - c^{2} = \frac{σ^{2}}{n} .

Finding the bias of $T_{2}$ , we need to find $E (T_{2}) - c^{2}$ .

E (T_{2}) = E (\frac{1}{n} n \sum j = 1 X_{j}^{2})

By linearity of expectation,

\begin{matrix} E (T_{2}) & = \frac{1}{n} n \sum j = 1 E (X_{j}^{2}) = \frac{1}{n} n \sum j = 1 (Var (X_{j}) + {(E (X_{j}))}_{j}^{2}) = \frac{1}{n} n \sum j = 1 (σ^{2} + c^{2}) = \frac{1}{n} \cdot n (σ^{2} + c^{2}) = σ^{2} + c^{2} . \end{matrix}

Hence the bias of $T_{2}$ is

σ^{2} + c^{2} - c^{2} = σ^{2} .

In fact, $T_{1}$ is the better estimator, as when $n$ becomes large, the bias of $T_{1}$ decreases, whereas the bias of $T_{2}$ does not.

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