Many LLM APIs give top-\(k\)
logprobs in their outputs. What if we want to obtain *all* the
logprobs? Here I present two algorithms for obtaining logprobs from an
LLM API. Both of these depend on the API allowing us to add a logit bias
to any token. The second is a generalization of the first.

Suppose for every call to the API we add a bias term \(b\in\mathbb{R}\) to a token \(i\in\mathbb{N}\). This means that the
model’s logits \(\boldsymbol\ell\) are
modified to \[\begin{aligned}
\boldsymbol\ell'=(\ell_1,\ell_2,\ldots,\ell_i+b,\ldots,\ell_v).
\end{aligned}\] If we collect the biased output \(\log
p_i'=\log\mathrm{softmax}(\boldsymbol\ell')_i\) for each
token \(i\), we now have \[\begin{aligned}
\log p_i'
&=\log\frac{\exp(\ell_i+b)}{\exp(\ell_i+b)+\sum_{j\neq i}\exp
\ell_j} \\
&= \log\frac{\exp\ell_i}{\exp\ell_i+\exp(-b)\sum_{j\neq i}\exp
\ell_j},
\end{aligned}\] which we can exponentiate and rearrange to get
\[\begin{aligned}
\frac{\exp(-b) p'_i}{1-p'_i} &=
\frac{\exp\ell_i}{\sum_{j\neq i}\exp\ell_j}.
\end{aligned}\] Note that the righthand side is the *odds*
of the token, therefore we can solve for the unbiased probability \(p_i\) of the token \[\begin{aligned}
\frac{p_i}{1-p_i} &= \frac{\exp(-b) p'_i}{1-p'_i} \\
p_i &= \frac{\exp(-b) p'_i}{1-p'_i+\exp(-b) p'_i} \\
\log p_i &= \log p'_i + \log
\frac{\exp(-b)}{1-p'_i+\exp(-b) p'_i} \\
\log p_i &= \log p'_i - \log\left(\exp b -\exp(b + \log
p'_i)+p'_i\right)
\label{eq:prob}
\end{aligned}\] Thus, it is possible to obtain unbiased logprobs
for any token with exactly 1 API call.

We can adapt this algorithm to solve for \(k\) logprobs at a time simultaneously. Without loss of generality, let us solve for tokens 1-\(k\) and assume that \(\exp\ell_1=1\) (since logprobs are invariant to uniform logit translation). Setting \(\lambda=\exp(-b)\) for convenience, we can solve for each \(\exp\ell_i\) \[\begin{aligned} p_i' &=\frac{\exp\ell_i}{\sum_{j\in[k]}\exp\ell_j+\lambda\sum_{j\not\in[k]}\exp\ell_j} \\ \frac{\exp\ell_i}{p_i'} &= \sum_{j\in[k]}\exp\ell_j + \lambda\sum_{j\not\in[k]}\exp\ell_j \\ \exp\ell_i &= \frac{p'_i}{p'_1} \end{aligned}\] which in turn allows us to solve for \(p_i\) \[\begin{aligned} \frac{\exp\ell_i}{p_i'} &= \sum_{j\in[k]}\exp\ell_j + \lambda\sum_{j\not\in[k]}\exp\ell_j \\ \frac{\exp\ell_i}{\lambda p_i'} - \frac1\lambda\sum_{j\in[k]}\exp\ell_j &= \sum_{j\not\in[k]}\exp\ell_j \\ \frac{\exp\ell_i}{\lambda p_i'} - \frac1\lambda\sum_{j\in[k]}\exp\ell_j &= \frac{\exp\ell_i}{p_i} - \sum_{j\in[k]}\exp\ell_j \\ p_i &= \frac{\lambda p'_i\exp\ell_i}{\exp\ell_i + p'_i(\lambda -1) \sum_{j\in[k]}\exp\ell_j} \\ p_i &= \frac{\lambda p'_i}{1 - \sum_{j\in[k]}p'_j + \lambda\sum_{j\in[k]}p'_j} \end{aligned}\] which we can take the log of an obtain \[\begin{aligned} \log p_i &= \log p'_i - \log\left(\left(1 - \sum_{j\in[k]}p'_j\right)\exp b + \sum_{j\in[k]}p'_j\right). \label{eq:parallel} \end{aligned}\]

Since most of the probability is often concentrated in the top-\(n\) tokens, one may wish to gather logprobs for only these top tokens and skip lower-probability tokens. This can be done by adding a penalty bias \(-b\) to tokens that have known logprob to surface the next-highest-probability batch of \(k\) tokens. We can then solve for the unbiased probability of new batch as follows. First, solve the biased softmax for \(\exp\ell_i\) \[\begin{aligned} p'_i &= \frac{\exp\ell_i}{\sum_{j=i}^v\exp\ell_j + \exp(-b)\sum_{j=1}^{i-1}\exp\ell_j} & \text{Biased softmax} \\ \exp\ell_i &= p'_i\left(\sum_{j=i}^v\exp\ell_j + \exp(-b)\sum_{j=1}^{i-1}\exp\ell_j\right) \label{eq:bs} \end{aligned}\] as well as the unbiased softmax \[\begin{aligned} p_i &= \frac{\exp\ell_i}{\sum_{j=1}^v\exp\ell_j} & \text{Unbiased softmax} \\ \exp\ell_i &= p_i\left(\sum_{j=1}^v\exp\ell_j\right) \label{eq:ubs} \end{aligned}\] which we can combine and simplify using the arbitrary assumption that \(\sum_{j=1}^{i-1}\exp\ell_j=1\) \[\begin{aligned} p_i\left(\sum_{j=1}^v\exp\ell_j\right) &= p'_i\left(\sum_{j=i}^v\exp\ell_j + \exp(-b)\sum_{j=1}^{i-1}\exp\ell_j\right) \\ p_i\left(1 + \sum_{j=i}^v\exp\ell_j\right) &= p'_i\left(\sum_{j=i}^v\exp\ell_j + \exp(-b)\right) & \text{Assume $\sum_{j=1}^{i-1}\exp\ell_j=1$} \label{eq:pi}. \end{aligned}\] We can now use the fact that we know the sum of the first \(i-1\) probabilities to solve for \(\sum_{j=i}^v\exp\ell_j\) \[\begin{aligned} \sum_{j=1}^{i-1}p_j&=\frac{\sum_{j=1}^{i-1}\exp\ell_j}{\sum_{j=1}^v\exp\ell_j} & \text{Softmax defn.} \\ &= \frac1{1+\sum_{j=i}^v\exp\ell_j} & \text{Recall $\sum_{j=1}^{i-1}\exp\ell_j=1$} \\ \sum_{j=i}^v\exp\ell_j &= \frac1{\sum_{j=1}^{i-1}p_j} - 1 \label{eq:sump} \end{aligned}\] and substitute this into our result from earlier to solve for \(p_i\) \[\begin{aligned} \frac{p_i}{\sum_{j=1}^{i-1}p_j} &= p'_i\left(\frac1{\sum_{j=1}^{i-1}p_j} - 1 + \exp(-b)\right) \\ p_i &= p'_i\left(1 + (\exp(-b)-1) \sum_{j=1}^{i-1}p_j\right). \end{aligned}\]