How to compute log likelihood for impossible events?

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I am defining a set $\mathbf{Z} = [p,q,r,s]$ such that $Pr(p)+Pr(q)+Pr(r)+Pr(s)=1$.

Likelihoods are defined as follows

\begin{align} \lambda_p&=\log \frac{Pr(x=p)}{Pr(x=s)}, \hspace{2mm} \lambda_q=\log \frac{Pr(x=q)}{Pr(x=s)}\\[15pt] \lambda_r&=\log \frac{Pr(x=r)}{Pr(x=s)}, \hspace{2mm} \lambda_s=\log \frac{Pr(x=s)}{Pr(x=s)}=0 \end{align}

If $Pr(p)=1$ then $Pr(q)=Pr(r)=Pr(s)=0$. In this case

\begin{align} \lambda_p&=\log \frac{Pr(x=p)}{Pr(x=s)}=\infty \end{align}

But I don't understand what values will $\lambda_q,\lambda_r,\lambda_s$ take.

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Your definition of log-likelihood is not standard. It is usually defined as the log of the probability of a given event given the distribution.

\begin{align} \lambda_p&=\log Pr(x=p), \hspace{2mm} \lambda_q=\log Pr(x=q) \dots \end{align}

it is always negative and is set to $-\infty$ when the event is impossible.

If you need to keep the quantities that you have defined, you have, actually, differences between log-likelihoods and they are not well defined for 2 likelihoods equal to $-\infty$

(I would have post that as a comment but do not have enough reputation for that)