Question

An individual whose level of exposure to a certain pathogen is \(x\) will contract the disease caused by this pathogen with probability \(P(x) .\) If the exposure level of a randomly chosen member of the population has probability density function \(f\), determine the conditional probability density of the exposure level of that member given that he or she (a) has the disease. (b) does not have the disease. (c) Show that when \(P(x)\) increases in \(x\), then the ratio of the density of part (a) to that of part (b) also increases in \(x\).

Short answer

In conclusion, we found the conditional probability density of the exposure level of a member given that he or she has the disease as: \(P_{A|B}(x) = \frac{P(x) \cdot f(x)}{\int P(x) \cdot f(x) dx}\) And the conditional probability density of the exposure level of a member given that he or she does not have the disease as: \(P_{A|B^c}(x) = \frac{[1 - P(x)] \cdot f(x)}{\int [1 - P(x)] \cdot f(x) dx}\) We also showed that when \(P(x)\) increases in \(x\), the ratio \(\frac{P_{A|B}(x)}{P_{A|B^c}(x)}\) also increases in \(x\).

Step-by-step solution

Recall Bayes' theorem for continuous random variables

The formula for Bayes' theorem in the continuous case is: \(P_{A|B}(x) = \frac{P_{B|A}(x) \cdot P_A(x)}{P_B(x)}\) We are given \(P_B(x)\) as probability of contracting the disease, so \(P_B(x) = P(x)\) and \(P_A(x)\) as exposure level probability density function, so \(P_A(x) = f(x)\). We need to find \(P_{A|B}(x)\), which is the probability density function of exposure level given the person has the disease.

Calculate the probability density function of having the disease given a certain exposure level

For this, we can use Bayes' theorem formula as follows: \(P_{A|B}(x) = \frac{P(x) \cdot f(x)}{P_B(x)}\) Since we need to find the probability density of the exposure level given that the individual has the disease, we need to find \(P_B(x)\),conditional probability density that a person has the disease given the density function \(f(x)\). This is calculated as: \(P_B(x) = \int P(x) \cdot f(x) dx\) Thus, the probability density function of exposure level given the individual has the disease is: \(P_{A|B}(x) = \frac{P(x) \cdot f(x)}{\int P(x) \cdot f(x) dx}\) (b) Find the conditional probability density for exposure level given the individual does not have the disease.

Calculate the probability density function of having the disease given a certain exposure level

First, we need to find the probability of not having the disease given a certain exposure level: \(P_{B^c}(x) = 1 - P(x)\) Now, we can use the Bayes' theorem formula again but this time to find the probability density function of exposure level given that the person does not have the disease. This will yield: \(P_{A|B^c}(x) = \frac{[1 - P(x)] \cdot f(x)}{P_{B^c}(x)}\) The conditional probability density that a person does not have the disease given the density function \(f(x)\) can be calculated as: \(P_{B^c}(x) = \int [1 - P(x)] \cdot f(x) dx\) Thus, the probability density function of exposure level given the individual does not have the disease is: \(P_{A|B^c}(x) = \frac{[1 - P(x)] \cdot f(x)}{\int [1 - P(x)] \cdot f(x) dx}\) (c) Prove that if \(P(x)\) increases with \(x\), the ratio of the density from part (a) to that of part (b) also increases with \(x\).

Analyze the ratio of probability densities

We want to show that when \(P(x)\) increases in \(x\), the ratio \( \frac{P_{A|B}(x)}{P_{A|B^c}(x)}\) also increases in \(x\). The ratio is given by: \( \frac{P_{A|B}(x)}{P_{A|B^c}(x)} = \frac{\frac{P(x) \cdot f(x)}{\int P(x) \cdot f(x) dx}}{\frac{[1 - P(x)] \cdot f(x)}{\int [1 - P(x)] \cdot f(x) dx}}\) \( = \frac{P(x) \cdot [\int [1 - P(x)] \cdot f(x) dx]}{[1 - P(x)] \cdot [\int P(x) \cdot f(x) dx]}\) Let's take the derivative of this ratio with respect to \(x\). If the derivative is positive, it means the ratio increases with \(x\): \(\frac{d}{dx} \left[\frac{P(x) \cdot [\int [1 - P(x)] \cdot f(x) dx]}{[1 - P(x)] \cdot [\int P(x) \cdot f(x) dx]}\right] > 0\) By analyzing this derivative, if it is always positive, it means the ratio increases when \(P(x)\) increases in \(x\).