Benchmarks

have more extraneous words. Therefore, we assume that

$$\backslash mathrm\{Pr\}(\backslash phi\_0|\backslash phi\_1^l,\; \backslash mathbf\{e\})\; =\; \backslash binom\{\backslash phi\_1\; +\; \backslash cdots\; +\; \backslash phi\_l\}\{\backslash phi\_0\}\; p\_0^\{\backslash phi\_1\; +\; \backslash cdots\; +\; \backslash phi\_l\; -\; \backslash phi\_0\}\; p\_1^\{\backslash phi\_0\}\; \backslash quad\; (1.30)$$

for some pair of auxiliary parameters $p\_0$ and $p\_1$ . The expression on the left-hand side of this equation depends on $\backslash phi\_1^l$ only through the sum $\backslash phi\_1\; +\; \backslash cdots\; +\; \backslash phi\_l$ and defines a probability distribution over $\backslash phi\_0$ whenever $p\_0$ and $p\_1$ are nonnegative and sum to 1. We can interpret $\backslash mathrm\{Pr\}(\backslash phi\_0|\backslash phi\_1^l,\; \backslash mathbf\{e\})$ as follows. We imagine that each of the words from $\backslash tau\_1^l$ requires an extraneous word with probability $p\_1$ and that this extraneous word must be connected to the empty cept. The probability that exactly $\backslash phi\_0$ of the words from $\backslash tau\_1^l$ will require an extraneous word is just the expression given in Equation (1.30).

As with Models 1 and 2, an alignment of $(\backslash mathbf\{f\}|\backslash mathbf\{e\})$ is determined by specifying $a\_j$ for each position in the French string. The fertilities, $\backslash phi\_0$ through $\backslash phi\_l$ , are functions of the $a\_j$ 's: $\backslash phi\_i$ is equal to the number of $j$ 's for which $a\_j$ equals $i$ . Therefore,

with $\backslash sum\_f\; t(f|e)\; =\; 1$ , $\backslash sum\_j\; d(j|i,\; m,\; l)\; =\; 1$ , $\backslash sum\_\backslash phi\; n(\backslash phi|e)\; =\; 1$ , and $p\_0\; +\; p\_1\; =\; 1$ . The assumptions that we make for Model 3 are such that each of the pairs $(\backslash tau,\; \backslash pi)$ in $\backslash langle\; \backslash mathbf\{f\},\; \backslash mathbf\{a\}\; \backslash rangle$ makes an identical contribution to the sum in Equation (1.29). The factorials in Equation (1.31) come from carrying out this sum explicitly. There is no factorial for the empty cept because it is exactly cancelled by the contribution from the distortion probabilities.

By now, the reader will be able to provide his own auxiliary function for seeking a constrained minimum of the likelihood of a translation with Model 3, but for completeness and to establish notation, we write

Following the trail blazed with Models 1 and 2, we define the counts

$$c(f|e;\; \backslash mathbf\{f\},\; \backslash mathbf\{e\})\; =\; \backslash sum\_\{\backslash mathbf\{a\}\}\; \backslash mathrm\{Pr\}(\backslash mathbf\{a\}|\backslash mathbf\{e\},\; \backslash mathbf\{f\})\; \backslash sum\_\{j=1\}^m\; \backslash delta(f,\; f\_j)\; \backslash delta(e,\; e\_\{a\_j\}),\; \backslash quad\; (1.33)$$

$$c(j|i,\; m,\; l;\; \backslash mathbf\{f\},\; \backslash mathbf\{e\})\; =\; \backslash sum\_\{\backslash mathbf\{a\}\}\; \backslash mathrm\{Pr\}(\backslash mathbf\{a\}|\backslash mathbf\{e\},\; \backslash mathbf\{f\})\; \backslash delta(i,\; a\_j),\; \backslash quad\; (1.34)$$

$$c(\backslash phi|e;\; \backslash mathbf\{f\},\; \backslash mathbf\{e\})\; =\; \backslash sum\_\{\backslash mathbf\{a\}\}\; \backslash mathrm\{Pr\}(\backslash mathbf\{a\}|\backslash mathbf\{e\},\; \backslash mathbf\{f\})\; \backslash sum\_\{i=1\}^l\; \backslash delta(\backslash phi,\; \backslash phi\_i)\; \backslash delta(e,\; e\_i),\; \backslash quad\; (1.35)$$

$$c(0;\; \backslash mathbf\{f\},\; \backslash mathbf\{e\})\; =\; \backslash sum\_\{\backslash mathbf\{a\}\}\; \backslash mathrm\{Pr\}(\backslash mathbf\{a\}|\backslash mathbf\{e\},\; \backslash mathbf\{f\})\; (m\; -\; 2\backslash phi\_0)$$