2
$\begingroup$

I have the following problem: Suppose we are interested in estimating the distribution over the English letters. We assume an alphabet that consists of 26 letters and the space symbol, and we ignore all other punctuation and the upper/lower case disctinction. We model the distribution over the 27 symbols as a multinomial parametrized by $\theta =(\theta_1,...,\theta_{27})$ where $\sum_i \theta_i = 1$ and all $\theta_i \geq 0$/

Now we go to Stanford's Green Library and repeat the following experiment: randomly pick up a book, open a page, pick a spot on the page, and write down the nearest symbol that is in our alphabet. We use $X[m]$ to denote the letter we obtain in the $mth$ experiment.

In the end we have collected a databse $D=\{x[1],...,x[2000] \}$ consisting of 2000 symbols, among which "a" appears 100times and "p" 87 times. We use a Dirichlet prior over $\theta$, i.e. $P(\theta)=Dirichlet(\alpha_1,...,\alpha_{27})$, where each $\alpha_i=10$.

Suppose we draw two more samples, X[2001] and X[2002]. If we use $\alpha_i=10$ for all $i$, what is the probability of $P(X[2001]="p",X[2002]="a"|D)$?

I thought we could compute: $P(x[2001]=p|D) \times P(x[2002]=a|D)=\frac{10+87}{270+2001} \times \frac{10+100}{270+2002}$ but it's wrong.

Recall formula : $P(x|u,D)=\frac{\alpha_{x,u}+M[x,u]}{\alpha_u+M[u]}$

$\endgroup$
1
$\begingroup$

Your formula is correct, but the final computing is wrong. It should be: $\frac{10+87}{270+2000} \times \frac{10+100}{270+2001}$

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.