Question

Let X be a random variable denoting the amount of loss in a business. The moment generating function of X is
\(M(t)=(\frac{2}{2-t})^2,\ \ \ \ t< 2.\)
If an insurance policy pays 60% of the loss, then the variance of the amount paid by the insurance policy equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 0.17

To find the variance of the amount a policyholder receives from insurance, given that the moment-generating function (MGF) of the random variable \(X\) is \(M(t)=\left(\frac{2}{2-t}\right)^2,\ t<2\) 

Let's break down the steps: 

1. **Determine the Mean and Variance of X:** The mean (expected value) and variance of a random variable can be derived using the MGF. The mean \(\mu\) is given by the first derivative of the MGF at \(t=0\), and the variance \(\sigma^2\) by the second derivative minus the square of the first derivative. Start with the MGF: \(M(t)=\left(\frac{2}{2-t}\right)^2\). The first derivative: \(M'(t)=2\left(\frac{2}{2-t}\right)\left(\frac{2}{(2-t)^2}\right)=\frac{8}{(2-t)^3}.\) Evaluate at \(t=0\) to get \(\mu:\) \(M'(0)=\frac{8}{8}=1.\) The second derivative: \(M''(t)=8\left(\frac{2}{2-t}\right)\left(\frac{1}{(2-t)^3}\right)=\frac{16}{(2-t)^4}.\) Evaluate at \(t=0\) to get \(M''(0)=1.\5.\) | Variance \(\sigma^2 = M''(0) - (M'(0))^2 = 1.5 - 1^2 = 0.5.\) 

2. **Calculate the Variance of the Insurance Payment:** The insurance policy pays 60% of the loss, so the payment \(Y=0.6X.\) Variance of \(Y\), \(\text{Var}(Y)=0.6^2\cdot\text{Var}(X)=0.36 \cdot 0.5 \ = 0.18.\) 

3. **Conclusion:** The variance of the amount paid by the insurance policy is \(0.18\), but slight rounding or tolerance issues may explain the tight range provided.