The Bernoulli distribution is based on the idea of carrying out a random experiment, which may result in a success or a failure. Let p be the probability of success; then, 1 − p is the probability of failure. If we assign the value 1 to variable X in case of success, and 0 otherwise, we get the following PMF:
It is easy to calculate the expected value:
and variance:
It is always useful to reflect a bit and get an intuitive feeling for the parameters of a probability distribution. If the support of the distribution is the set {0,1}, the expected value is the same as the parameter p. Hence, the expected value of X is large when the probability of success is large, and this makes obvious sense. The same applies if you generalize the support to any other value. As a practical example, consider the launch of a new product and the corresponding profit if it is a success or a failure. Having only two values would arguably make a poor model of uncertainty, but it may be a simple starting point.14
Checking variance is a bit more interesting and leads us to a few useful observations:
- Variance cannot be negative, since p ∈ [0, 1]. A formula like p(p − 1) would make no sense, since it would allow for a negative variance. Indeed, if we plot the formula for variance as a function of p, we get a concave parabola such that variance is positive for p in the allowable range, 0 ≤ p ≤ 1, and negative outside.
- Variance is zero, the minimum possible value, when p = 0 and p = 1. This makes sense again; if success or failure are guaranteed, there is no uncertainty.
- The maximum variance is obtained when p = 0.5, which corresponds to the maximum level of uncertainty about the outcomes.
Now let us use this intuition in a toy example that is related to the Bernoulli random variable.
Fig. 6.3 Partial decision tree in Example 6.10.
Example 6.10 A firm has developed a new product and it must decide whether starting full scale production is worthwhile.15 If the product is successful, profit will be €120,000; otherwise, it will just be €20,000. Probability of success is 0.6, or 60%, so there is quite some uncertainty. As an alternative, the firm could just sell the production license for €60,000 to another firm. However, there is still a third possibility; the firm could try to get some additional information by carrying out a market research survey, at a cost of €4,000. If the result of the survey is promising, the firm believes that the probability of success will be 0.9; otherwise, it will just be 0.3. Note that these are conditional probabilities, given the outcome of the survey.
- At what level of probability would the survey suggest a promising future for the product?
- What is the best course of action for the firm?
- What is the maximum cost of the survey that the firm should be willing to pay?
The first question seems a bit tricky. The point is that by carrying out a small-scale survey, interviewing potential customers to see if they like the new product, we will not increase the unconditional probability of success. The firm is not improving the product; it is only gathering information. The situation can be visualized by the tree in Fig. 6.3. The events we consider are POK, the product is successful; Sgood, the survey result is promising; Sbad, the survey result is disappointing. If we do not carry out the survey, unconditional probability of success is 60%. If we carry out the survey, the unconditional probability of success, as seen from the root of the tree, is
Setting this equal to 0.6 (60%) and solving for the unknown probability yields p = 0.5. This is the only probability ensuring consistency in our representation of uncertainty. Now, if we compare the expected profit from immediate product launch, we obtain
against the profit from selling a production license, we see that we should give the product a try.16 If we carry out the survey, and the result is positive, expected profit is
from which the cost of the survey should be deducted. If the survey is not promising, then
and we are better off selling the license. Hence, if we carry out the survey and make optimal use of the information that it provides, we obtain
We see that the best course of action is:
- Carrying out the survey.
- If the result is promising, start production.
- If the result is not promising, sell the license.
Of course this is just a toy example, but it is important to really understand what does the trick. The survey does not improve our chances of success. Rather, it allows us to defer the decision while collecting additional information. We may see this by recalling what we have just observed about the Bernoulli random variable: Variance is decreased when the probability of success is driven away from 0.5, one way or the other. In this example, the unconditional standard deviation of profit if we do not carry out the survey will be
The conditional standard deviation depending on the survey outcome are
We see that the effect of the survey is to reduce uncertainty, and the firm should be willing to pay at most €5000 for this reduction. To see this, observe that the expected profit in Eq. (6.14), without deducting the cost of the survey, is €85,000, which should be compared against €80,000, the expected profit of immediate product launch from Eq. (6.13).
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