统计代写|贝叶斯统计代写Bayesian statistics代考|EXAMPLES: MULTINOMIAL MODELS

Categorized data, such as one encounters frequently in genetics (where the categories are usually phenolypes), give rise to multinomial models. Since the distribution theory of multinomial models is well known and especially elegant, they will serve the illustrative purposes of this section well. Some slight familiarity with Mendelian genetics will be assumed. ${ }^4$

EXAMPLE 1 (genetic linkage). Two factors are ‘linked’ when they lie on the same chromosome. As is customary, we denote the dominant alleles by capital letters, and reserve small letters for the recessive alleles. The heterozygote can occur in either of two phases: coupling $(A B a b)$ where $A$ and $B$ lie on one chromosome of a homologous pair and $a, b$ on the other, or repulsion $(A b a B)$, where a dominant $A$ is linked to a recessive $b$ and vice versa. Chromosomes may break and members of a pair sometimes exchange homologous sections. As a result, the two chromosomes $A b$ and $a B$ of repulsion phase may give rise to the pair $A B, a b$ of coupling phase (or vice versa). The rejoined chromosomes are called recombinants and crossing-over is said to have occurred.

Let $\theta$ be the probability of crossing-over (called the ‘recombination fraction’). Then the gametic output of the heterozygote $A b a B$ will be $A B$, $A b, a B$, and $a b$ in the relative frequencies:
$$\text { (5.4) } \quad \theta / 2,(1-\theta) / 2,(1-\theta) / 2, \theta / 2 \text {. }$$
For crossing-over occurs with probability $\theta$, and then one member of each chromosome pair is $A B$ and the other is $a b$, and similarly for the non-recombinant types $A b$ and $a B$. If we cross the heterozygote to a double recessive, $a b a b$, a so-called ‘double backcross’, the four phenotypes $A b, A b$, $a B$, and $a b$ will also be genotypes, since the recessive genes from the doubly recessive parent have nil effect. Hence, (5.4) gives the category probabilities for the double backcross, the categories being the phenotypes (and genotypes) $A B, A b, a B, a b$. Let $n$ be the sample size, and let $n_i$ be the observed number of off-spring in category $i, i=1,2,3,4$.

统计代写|贝叶斯统计代写Bayesian statistics代考|MULTINOMIAL MODELS WITHOUT FREE PARAMETERS

Let the random vector $\left(X_1, \ldots, X_k\right)$ have the multinomial distribution with $k$ categories and category probabilities $p_1, \ldots, p_k$ :
$$P\left(X_1=x_1, \ldots, X_k=x_k\right)=\frac{n !}{x_{1} ! \ldots x_{k-1} ! x_{k} !} p_1^{x_1} \ldots p_k^{x_k}$$
where $x_i$ is the observed number in cateogry $i$ and $\Sigma x_i=n$, the sample size.

Then the vector $\left(X_1, \ldots, X_{k-1}\right)$ has, asymptotically, a non-singular normal distribution with density
$(5.12)(2 \pi)^{-(k-1) / 2} \Lambda^{-1 / 2} e^{-Q / 2}$,
where
$$Q=\sum_{i, j}^{k-1}\left(\Lambda_{i j} / \Lambda\right)\left(X_i-n p_i\right)\left(X_j-n p_j\right)$$
is the quadratic form whose matrix of coefficients is inverse to the covariance matrix of the multinomial with diagonal entries $n p_i\left(1-p_i\right)$ and off-diagonal entries $-n p_i p_j, i \neq j$. The $\Lambda_{i j}$ are thus the co-factors of the covariance matrix and $\Lambda$ the determinant. Given the obvious analogy to the univariate case, $\Lambda$ is called the generalized variance of the distribution and its square root the generalized standard deviation. It is well known that $Q$ has a chi square distribution with $k-1$ degrees of freedom. Hence
(5.13) $Q=\chi_{\alpha, k-1}^2$
is the equation of a $100(1-q) \%$ direct confidence ellipsoid for the vector $\left(X_1, \ldots, X_{k-1}\right)$, writing $\chi_{\alpha, k-1}^2$ for the upper $100 \alpha \%$ point of the chi square distribution with $k-1 d f$. (More generally, the experimental outcome will fall in a $100(1-\alpha) \%$ direct confidence region (or $D C R$ ) with probability $100(1-\alpha)$, it being assumed that the probability of every included outcome, conditional on the model, exceeds that of any excluded outcome.)

贝叶斯统计代考

统计代写|贝叶斯统计代写Bayesian statistics代考|EXAMPLES: MULTINOMIAL MODELS

(5.4) $\theta / 2,(1-\theta) / 2,(1-\theta) / 2, \theta / 2$.

统计代写|贝叶斯统计代写Bayesian statistics代考|MULTINOMIAL MODELS WITHOUT FREE PARAMETERS

$$P\left(X_1=x_1, \ldots, X_k=x_k\right)=\frac{n !}{x_{1} ! \ldots x_{k-1} ! x_{k} !} p_1^{x_1} \ldots p_k^{x_k}$$

$$(5.12)(2 \pi)^{-(k-1) / 2} \Lambda^{-1 / 2} e^{-Q / 2} \text { ， }$$

$$Q=\sum_{i, j}^{k-1}\left(\Lambda_{i j} / \Lambda\right)\left(X_i-n p_i\right)\left(X_j-n p_j\right)$$

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