Question

Consider a random variable t∼Bernoulli(π) for π<0.5. For example, Pr[t=1]=π,Pr[t=0]=1−π. Suppose the prediction also follows a Bernoulli distribution: t^∼Bernoulli(p) Compute the expected accuracy E[I{t=t^}], where I{⋅} is an indicator function. What is the value of p that yields the highest expected accuracy?

Answer 1

To find the value of p that yields the highest expected accuracy in a Bernoulli distribution problem, we can compute the expected accuracy E[I{t=t^}] and then take the derivative of E[I{t=t^}] with respect to p, setting it equal to zero.

Step-by-step explanation:

In this problem, we are given that the random variable t follows a Bernoulli distribution with parameter π, where π < 0.5. We are also given that the prediction t^ also follows a Bernoulli distribution with parameter p. We need to compute the expected accuracy E[I{t=t^}] and determine the value of p that yields the highest expected accuracy.

The expected accuracy E[I{t=t^}] is simply the probability that t and t^ are equal. Since t and t^ both follow Bernoulli distributions, we can express this probability as Pr[t=1 and t^=1] + Pr[t=0 and t^=0]. Substituting the given parameters π and p, we get E[I{t=t^}] = πp + (1-π)(1-p) = p + π - 2πp.

To find the value of p that yields the highest expected accuracy, we can take the derivative of E[I{t=t^}] with respect to p and set it equal to zero. Solving this equation will give us the value of p that maximizes the expected accuracy.