The Value of Information

Should you invest or not? A signal helps you decide.

The more accurate the signal, the better the decision. After seeing the signal, you recompute the probability. A good signal says invest. A bad signal says stop. The value of information is the gap between deciding with it and without it. The formula below pins that gap down.

For example, before buying a used πŸš— car, you ask an expert. The expert spots good and bad cars correctly 80 percent of the time. If the expert says good, you buy. If the expert says bad, you walk away. Your losses shrink a lot.

P(good∣h)=Ο€qΟ€q+(1βˆ’Ο€)(1βˆ’q)P(\text{good} \mid h) = \frac{\pi q}{\pi q + (1-\pi)(1-q)}
VoI=P(h)max⁑(0,EVh)+P(l)max⁑(0,EVl)βˆ’max⁑(0,Ο€Rβˆ’(1βˆ’Ο€)L)\text{VoI} = P(h)\max(0, \text{EV}_h) + P(l)\max(0, \text{EV}_l) - \max(0, \pi R - (1-\pi)L)

Part of these learning paths

Examples & Exercises

Let Ο€=1/2\pi = 1/2, signal accuracy q=0.8q = 0.8, R=100R = 100, L=60L = 60.

Compute by hand the optimal expected payoff without a signal, the posterior and optimal action after each signal realization, and the value of information (VoI).

EVno=max⁑(0,Ο€Rβˆ’(1βˆ’Ο€)L),VoI=EVwithβˆ’EVno\text{EV}_{\text{no}} = \max(0, \pi R - (1-\pi)L), \qquad \text{VoI} = \text{EV}_{\text{with}} - \text{EV}_{\text{no}}
P(good∣h)=Ο€qΟ€q+(1βˆ’Ο€)(1βˆ’q),P(good∣l)=Ο€(1βˆ’q)Ο€(1βˆ’q)+(1βˆ’Ο€)qP(\text{good} \mid h) = \frac{\pi q}{\pi q + (1-\pi)(1-q)}, \qquad P(\text{good} \mid l) = \frac{\pi(1-q)}{\pi(1-q) + (1-\pi)q}
Ο€=1/2\pi = 1/2, q=0.8q = 0.8, R=100R = 100, L=60L = 60 β€” VoI = 14.
Show proof

  1. The posteriors are P(good∣h)=0.4/0.5=0.8P(\text{good} \mid h) = 0.4/0.5 = 0.8 and P(good∣l)=0.1/0.5=0.2P(\text{good} \mid l) = 0.1/0.5 = 0.2. Without a signal, Ο€Rβˆ’(1βˆ’Ο€)L=50βˆ’30=20>0\pi R - (1-\pi)L = 50 - 30 = 20 > 0, so invest and EVno=20\text{EV}_{\text{no}} = 20.

  2. After hh, 0.8β‹…100βˆ’0.2β‹…60=68>00.8 \cdot 100 - 0.2 \cdot 60 = 68 > 0, so invest; after ll, 0.2β‹…100βˆ’0.8β‹…60=βˆ’28<00.2 \cdot 100 - 0.8 \cdot 60 = -28 < 0, so stay out. With P(h)=0.5P(h) = 0.5, EVwith=0.5β‹…68+0.5β‹…0=34\text{EV}_{\text{with}} = 0.5 \cdot 68 + 0.5 \cdot 0 = 34.

  3. Hence VoI=34βˆ’20=14\text{VoI} = 34 - 20 = 14. The signal's value comes entirely from avoiding the loss after ll β€” after good news the investor would have invested anyway.

    VoI=0.5β‹…68⏟afterΒ h+0.5β‹…0⏟afterΒ lβˆ’20=14\text{VoI} = \underbrace{0.5 \cdot 68}_{\text{after } h} + \underbrace{0.5 \cdot 0}_{\text{after } l} - 20 = 14

Let Ο€=0.2\pi = 0.2, q=0.8q = 0.8, R=100R = 100, L=60L = 60.

Find the optimal action without a signal, show the signal can reverse that decision, and compute the VoI.

Ο€Rβˆ’(1βˆ’Ο€)L=20βˆ’48=βˆ’28<0\pi R - (1-\pi)L = 20 - 48 = -28 < 0
P(good∣h)=0.160.16+0.16=0.5P(\text{good} \mid h) = \frac{0.16}{0.16 + 0.16} = 0.5
Ο€=0.2\pi = 0.2, q=0.8q = 0.8, R=100R = 100, L=60L = 60 β€” the signal reverses the decision.
Show proof

  1. Without a signal, Ο€Rβˆ’(1βˆ’Ο€)L=20βˆ’48=βˆ’28<0\pi R - (1-\pi)L = 20 - 48 = -28 < 0, so staying out is optimal and EVno=0\text{EV}_{\text{no}} = 0.

  2. Since P(good∣h)=0.16/(0.16+0.16)=0.5P(\text{good} \mid h) = 0.16/(0.16 + 0.16) = 0.5, the expected payoff of invest after hh is 0.5β‹…100βˆ’0.5β‹…60=20>00.5 \cdot 100 - 0.5 \cdot 60 = 20 > 0 β€” the decision flips to invest. After ll, P(good∣l)=0.04/0.68β‰ˆ0.0588P(\text{good} \mid l) = 0.04/0.68 \approx 0.0588 with expected payoff β‰ˆβˆ’50.59\approx -50.59, so staying out remains optimal.

  3. With P(h)=0.16+0.16=0.32P(h) = 0.16 + 0.16 = 0.32, EVwith=0.32β‹…20=6.4\text{EV}_{\text{with}} = 0.32 \cdot 20 = 6.4 and VoI=6.4βˆ’0=6.4\text{VoI} = 6.4 - 0 = 6.4. For someone who would never invest a priori, the signal has positive value as the instruction "sometimes invest".

    VoI=0.32β‹…20βˆ’0=6.4\text{VoI} = 0.32 \cdot 20 - 0 = 6.4

18. Economics of Information