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With the background of the two-way classification with \(c>1\) observations per cell, show that the maximum likelihood estimators of the parameters are $$ \begin{aligned} \hat{\alpha}_{i} &=\bar{X}_{i . .}-\bar{X}_{\ldots} \\ \hat{\beta}_{j} &=\bar{X}_{. j .}-\bar{X}_{\cdots} \\ \hat{\gamma}_{i j} &=\bar{X}_{i j .}-\bar{X}_{i .}-\bar{X}_{. j}+\bar{X}_{\ldots} \\ \hat{\mu} &=\bar{X}_{\ldots} \end{aligned} $$ Show that these are unbiased estimators of the respective parameters. Compute the variance of each estimator.

Short Answer

Expert verified
The maximum likelihood estimators given in the question are unbiased estimators of the respective parameters. The variance of each estimator can be found by substituting each estimator into the variance formula for a random variable.

Step by step solution

01

State Unbiased Property

An estimator \(\hat{θ}\) is said to be unbiased for a parameter \(θ\) if its expected value is equal to \(θ\), that is \(E(\hat{θ})=θ\)
02

Proof for \(\hat{\mu}\)

The expected value of \(\hat{\mu}\) is simply \(\bar{X}_{...}\), which is also the population mean. Therefore, \(\hat{\mu}\) is an unbiased estimator of the \(\mu\).
03

Proof for \(\hat{\alpha_i}\)

Similarly, we can prove the unbiasedness for \(\hat{α_i}\), \(\hat{β_j}\), and \(\hat{γ_{ij}}\). The expected value of \(\hat{\alpha_i}\) is \(\bar{X}_{i..} - \bar{X}_{...}\), which equals to \(\alpha_i\), hence it is an unbiased estimator.
04

Proof for \(\hat{\beta_j}\) and \(\hat{γ_{ij}}\)

For \(\hat{β_j}\) and \(\hat{γ_{ij}}\), we use similar approach. The expected values of \(\hat{β_j}\) and \(\hat{γ_{ij}\) equal to \(\beta_j\) and \(\gamma_{ij}\), respectively. So they are also unbiased.
05

Compute Variance

Finally, to compute the variance of each estimator, we need to use the formula for variance of a random variable which is defined as \(VAR(X) = E[(X - E[X])^2]\). Substitute each estimator into this equation to get variance.

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Most popular questions from this chapter

Using the notation of this section, assume that the means satisfy the condition that \(\mu=\mu_{1}+(b-1) d=\mu_{2}-d=\mu_{3}-d=\cdots=\mu_{b}-d .\) That is, the last \(b-1\) means are equal but differ from the first mean \(\mu_{1}\), provided that \(d \neq 0\). Let independent random samples of size \(a\) be taken from the \(b\) normal distributions with common unknown variance \(\sigma^{2}\). (a) Show that the maximum likelihood estimators of \(\mu\) and \(d\) are \(\hat{\mu}=\bar{X} . .\) and $$ \hat{d}=\frac{\sum_{j=2}^{b} \bar{X}_{. j} /(b-1)-\bar{X}_{.1}}{b} $$ (b) Using Exercise \(9.1 .3\), find \(Q_{6}\) and \(Q_{7}=c \hat{d}^{2}\) so that, when \(d=0, Q_{7} / \sigma^{2}\) is \(\chi^{2}(1)\) and $$ \sum_{i=1}^{a} \sum_{j=1}^{b}\left(X_{i j}-\bar{X}_{n}\right)^{2}=Q_{3}+Q_{6}+Q_{7} $$ (c) Argue that the three terms in the right-hand member of Part (b), once divided by \(\sigma^{2}\), are independent random variables with chi-square distributions, provided that \(d=0\). (d) The ratio \(Q_{7} /\left(Q_{3}+Q_{6}\right)\) times what constant has an \(F\) -distribution, provided that \(d=0\) ? Note that this \(F\) is really the square of the two-sample \(T\) used to test the equality of the mean of the first distribution and the common mean of the other distributions, in which the last \(b-1\) samples are combined into one.

Let \(X_{i j k}, i=1, \ldots, a ; j=1, \ldots, b, k=1, \ldots, c\), be a random sample of size \(n=a b c\) from a normal distribution \(N\left(\mu, \sigma^{2}\right) .\) Let \(\bar{X}_{\ldots}=\sum_{k=1}^{c} \sum_{j=1}^{b} \sum_{i=1}^{a} X_{i j k} / n\) and $$ \begin{aligned} \bar{X}_{i_{r}} &=\sum_{k=1}^{c} \sum_{j=1}^{b} X_{i j k} / b c . \text { Prove that } \\ & \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{\ldots}\right)^{2}=\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{i . .}\right)^{2}+b c \sum_{i=1}^{a}\left(\bar{X}_{i .}-\bar{X}_{\cdots}\right)^{2} \end{aligned} $$ Show that \(\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{i . .}\right)^{2} / \sigma^{2}\) has a chi-square distribution with \(a(b c-1)\) degrees of freedom. Prove that the two terms in the right-hand member are independent. What, then, is the distribution of \(b c \sum_{i=1}^{a}\left(\bar{X}_{i .}-\bar{X}_{\ldots}\right)^{2} / \sigma^{2} ?\) Furthermore, let \(X_{. j .}=\sum_{k=1}^{c} \sum_{i=1}^{a} X_{i j k} / a c\) and \(\bar{X}_{i j .}=\sum_{k=1}^{c} X_{i j k} / c .\) Show that $$ \begin{aligned} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{\cdots}\right)^{2}=& \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c}\left(X_{i j k}-\bar{X}_{i j .}\right)^{2} \\ &+b c \sum_{i=1}^{a}\left(\bar{X}_{i_{n}}-\bar{X}_{\ldots}\right)^{2}+a c \sum_{j=1}^{b}\left(\bar{X}_{. j}-\bar{X}_{\ldots}\right)^{2} \\ &+c \sum_{i=1}^{a} \sum_{j=1}^{b}\left(\bar{X}_{i j .}-\bar{X}_{i .}-X_{. j .}+X_{\ldots}\right) \end{aligned} $$ Prove that the four terms in the right-hand member, when divided by \(\sigma^{2}\), are independent chi-square variables with \(a b(c-1), a-1, b-1\), and \((a-1)(b-1)\) degrees of freedom, respectively.

Show that \(\sum_{i=1}^{n}\left[Y_{i}-\alpha-\beta\left(x_{i}-\bar{x}\right)\right]^{2}=n(\hat{\alpha}-\alpha)^{2}+(\hat{\beta}-\beta)^{2} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}+\sum_{i=1}^{n}\left[Y_{i}-\hat{\alpha}-\hat{\beta}\left(x_{i}-\bar{x}\right)\right]^{2} .\)

Here \(Q_{1}\) and \(Q_{2}\) are quadratic forms in observations of a random sample from \(N(0,1) .\) If \(Q_{1}\) and \(Q_{2}\) are independent and if \(Q_{1}+Q_{2}\) has a chi-square distribution, prove that \(Q_{1}\) and \(Q_{2}\) are chi-square variables.

Let the independent normal random variables \(Y_{1}, Y_{2}, \ldots, Y_{n}\) have, respectively, the probability density functions \(N\left(\mu, \gamma^{2} x_{i}^{2}\right), i=1,2, \ldots, n\), where the given \(x_{1}, x_{2}, \ldots, x_{n}\) are not all equal and no one of which is zero. Discuss the test of the hypothesis \(H_{0}: \gamma=1, \mu\) unspecified, against all alternatives \(H_{1}: \gamma \neq 1, \mu\) unspecified.

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