In this problem, we will model the likelihood of a particular client of a financial firm defaulting on his or her loans based on previous transactions. There are only two outcomes, “Yes" or “No", depending on whether the client eventually defaults or not. It is believed that the client's current balance is a good predictor for this outcome, so that the more money is spent without paying, the more likely it is for that person to default.
For each x, we will write Y_ x as the 0-1 outcome of defaulting/not defaulting, given a particular current balance x. In other words, we will model the distribution of Y_ x as a Bernoulli distribution, which is reasonable given that there are only two possible outcomes.
First, recall the likelihood of a Bernoulli RV Y in terms of the parameter p:
P(Y = y) = p^ y (1-p)^{1-y}.
Rewrite this in terms of an exponential family
P(Y = y) = h(y) \exp \left[ \eta (p) T(y) - B(p) \right].
Since this representation is only unique up to re-scaling by constants, take the convention that T(y) = y.
\eta (p) =
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B(p) =
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h(y) =
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We can write this in canonical form, e.g. as
P(Y = y) = h(y) \exp \left[ y \eta - b(\eta ) \right].
What is b(\eta )?
b(\eta ) =
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Recall that the mean of a Bernoulli(p) distribution is p. What is the canonical link function \, g(\mu ) \, associated with this exponential family, where \, \mu = \mathbb E[Y] \,? Write your answer in terms of p.
g(\mu ) =
Since the mean of a Bernoulli(p) distribution is p, the canonical link function g(\mu) is given by:
g(\mu) = log(\frac{\mu}{1-\mu}) = log(\frac{p}{1-p})
To find the canonical link function g(μ) associated with the exponential family in question, let's start by recalling that the mean μ of a Bernoulli(p) distribution is equal to p.
The canonical link function g(μ) is defined as the function that relates the linear predictor η to the mean μ. In this case, we have:
μ = p
To find g(μ), we need to solve for p in terms of μ:
p = μ
Therefore, the canonical link function g(μ) for this exponential family is simply:
g(μ) = μ