Break-Even Price of a Digital with Two Independent Signals
An event $E$ has prior probability $p$. Before trading, you observe two independent binary signals $S_1$ and $S_2$ with known conditional probabilities:
$P(S_1 = 1 \mid E) = t_1, \quad P(S_1 = 1 \mid E^c) = f_1$ $P(S_2 = 1 \mid E) = t_2, \quad P(S_2 = 1 \mid E^c) = f_2$
The signals are conditionally independent given $E$ (and given $E^c$). A digital contract pays $\
- Derive the posterior probability $p_{11} = P(E \mid S_1 = 1, S_2 = 1)$. State the analogous formula for $p_{00} = P(E \mid S_1 = 0, S_2 = 0)$.
- You can buy one unit of the digital at price $m$ when $(S_1, S_2) = (1,1)$, and sell one unit at price $m$ when $(S_1, S_2) = (0,0)$. Find the break-even $m$ for each case.
- Discuss which signal corner -- $(1,1)$ or $(0,0)$ -- is more informative, and how this depends on the likelihood ratios $t_i / f_i$.
Hints
- The fair price of a digital is the posterior probability of the payoff event. Focus on computing posteriors.
- Conditional independence means the joint likelihood factors: $P(S_1=1, S_2=1 \mid E) = P(S_1=1 \mid E) \cdot P(S_2=1 \mid E)$. In odds form, the likelihood ratios multiply.
- To compare informativeness of the $(1,1)$ and $(0,0)$ corners, compare the absolute log-likelihood ratios $|\ln(t_1 t_2 / f_1 f_2)|$ and $|\ln((1-t_1)(1-t_2) / (1-f_1)(1-f_2))|$.
Worked Solution
How to Think About It: You are a market maker pricing a digital contract after observing two noisy signals about the underlying event. The fair price of the digital is just the posterior probability of $E$ given the signals. Two confirming signals should move the price more than one -- the question is exactly how they combine. Since the signals are conditionally independent, their likelihood ratios multiply. This is the key: two independent pieces of evidence combine multiplicatively, not additively.
Quick Estimate: Suppose $p = 0.5$, $t_1 = 0.8$, $f_1 = 0.2$, $t_2 = 0.7$, $f_2 = 0.3$. The likelihood ratio for $S_1 = 1$ is $0.8/0.2 = 4$. For $S_2 = 1$, it is $0.7/0.3 \approx 2.33$. Combined: $4 \times 2.33 \approx 9.33$. With equal prior, posterior odds are $9.33:1$, so $p_{11} \approx 9.33/10.33 \approx 0.90$. For $(0,0)$: likelihood ratios are $(1-0.8)/(1-0.2) = 0.25$ and $(1-0.7)/(1-0.3) \approx 0.43$. Combined: $0.25 \times 0.43 \approx 0.107$. Posterior odds: $0.107:1$, so $p_{00} \approx 0.107/1.107 \approx 0.097$. The $(1,1)$ corner is very bullish; the $(0,0)$ corner is very bearish.
Approach: Apply Bayes' theorem with conditional independence to factor the joint likelihoods.
Formal Solution:
Part 1: Posterior $p_{11}$.
By Bayes' theorem:
$p_{11} = P(E \mid S_1=1, S_2=1) = \frac{P(S_1=1, S_2=1 \mid E) \, p}{P(S_1=1, S_2=1)}$
By conditional independence given $E$:
$P(S_1=1, S_2=1 \mid E) = t_1 \, t_2$ $P(S_1=1, S_2=1 \mid E^c) = f_1 \, f_2$
So the denominator is $t_1 t_2 \, p + f_1 f_2 \, (1-p)$, giving:
$p_{11} = \frac{t_1 \, t_2 \, p}{t_1 \, t_2 \, p + f_1 \, f_2 \, (1-p)}$
Equivalently, in odds form: the posterior odds equal the prior odds times the combined likelihood ratio:
$\frac{p_{11}}{1 - p_{11}} = \frac{p}{1-p} \cdot \frac{t_1}{f_1} \cdot \frac{t_2}{f_2}$
Analogously for $(0,0)$:
$p_{00} = \frac{(1-t_1)(1-t_2) \, p}{(1-t_1)(1-t_2) \, p + (1-f_1)(1-f_2) \, (1-p)}$
$\frac{p_{00}}{1 - p_{00}} = \frac{p}{1-p} \cdot \frac{1-t_1}{1-f_1} \cdot \frac{1-t_2}{1-f_2}$
Part 2: Break-even price $m$.
The digital pays $\
$m_{\text{buy}} = p_{11} = \frac{t_1 \, t_2 \, p}{t_1 \, t_2 \, p + f_1 \, f_2 \, (1-p)}$
When $(S_1, S_2) = (0,0)$, you sell at the break-even price:
$m_{\text{sell}} = p_{00} = \frac{(1-t_1)(1-t_2) \, p}{(1-t_1)(1-t_2) \, p + (1-f_1)(1-f_2) \, (1-p)}$
Part 3: Which corner is more informative?
The informativeness of a signal corner depends on how far it moves the posterior from the prior. This is governed by the likelihood ratios:
- For $(1,1)$: combined LR $= \frac{t_1 t_2}{f_1 f_2}$
- For $(0,0)$: combined LR $= \frac{(1-t_1)(1-t_2)}{(1-f_1)(1-f_2)}$
The corner with the larger absolute log-likelihood ratio is more informative (moves the posterior further from the prior). If the signals have high true-positive rates ($t_i$ large) and low false-positive rates ($f_i$ small), then $t_i/f_i$ is large and $(1,1)$ is very informative. Conversely, if the signals are better at ruling out $E$ (high
In general, it is not symmetric. For instance, if $t_1 = t_2 = 0.9$ and $f_1 = f_2 = 0.1$, the $(1,1)$ LR is $81$ and the $(0,0)$ LR is
Answer: $p_{11} = \frac{t_1 t_2 \, p}{t_1 t_2 \, p + f_1 f_2(1-p)}$ and $p_{00} = \frac{(1-t_1)(1-t_2)p}{(1-t_1)(1-t_2)p + (1-f_1)(1-f_2)(1-p)}$. The break-even prices equal these posteriors. The more informative corner has the larger absolute log-likelihood ratio; when $t_i + f_i = 1$, both corners carry the same information.
Intuition
This problem illustrates the multiplicative nature of Bayesian updating with independent evidence. Each signal contributes a likelihood ratio, and independent signals multiply. This is why even weak signals become powerful in combination -- two signals with likelihood ratios of 3 give a combined ratio of 9, which can move the posterior dramatically from the prior.
In market making, this is how you should think about combining information sources. If you observe order flow from two independent informed traders, their combined information is multiplicative in the odds, not additive. This means the bid-ask spread should widen more than linearly with the number of informed signals, and the optimal quote after seeing confirming signals moves much further than you might naively expect from just one signal.