[Math] Exponential distribution problem 2

exponential distributionprobability

A car is new at the beginning of a calendar year. The time, in years, before the car
experiences its first failure is exponentially distributed with mean 2.
Calculate the probability that the car experiences its first failure in the last quarter of
some calendar year.

Step 1:Let x=time in years before car experience first failure.

$$\sum \limits_{n=0}^{\infty} P(0.75+n<x<1+n)$$ n=some year

Step 2: Since x exponential the cumulative function is $F(x)=1-$e^-(x/$\theta$)

Step 3. $F(1)-F(0.75)=0.081$

the correct answer is 0.205,can someone help me get on the right track to solve this problem.

Best Answer

Your series summation results in

$$\frac{e^{-0.75/\theta}-e^{-1/\theta}}{1-e^{-1/\theta}}$$

$$\frac{F(1)-F(0,75)}{F(1)} = 0.205 \, .$$

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[Math] Actuarial Exam P problem (expected value of a max function.)

Comment. Here is a simulation in R statistical software with very nearly the correct numerical answer. A direct method based on $X$ is used. You can compare the result (to two or three places) with your analytic answer.

t = rexp(10^6, rate = 1/3)
x = pmax(x,2)
mean(x)       
## 3.538123          # aprx E(X) from simulation

mean(t > 2)
## 0.51334           # aprx P(T > 2) from simulation
1 - pexp(2, 1/3)
## 0.5134171         # exact P(T > 2)

A little over half of the devices survive beyond 2.

[Math] Finding probability that a car experiences a failure

I think you just misread the question, it means the first failure in the last quarter of some calendar year, it need not be the first year.

\begin{align} \sum_{n=0}^\infty P\left(n+1 > T > n + \frac34\right) &=\sum_{n=0}^\infty \left(F(n+1)-F\left(n+\frac34\right)\right)\\ &= \sum_{n=0}^\infty \left(\exp \left(-\frac{n+\frac34}2 \right) - \exp\left( -\frac{n+1}{2}\right) \right)\\ &= \left( \exp\left(-\frac38 \right) -\exp\left( -\frac12\right)\right)\sum_{n=0}^\infty \exp(-n/2) \\ &=\frac{\left( \exp\left(-\frac38 \right) -\exp\left( -\frac12\right)\right)}{1-\exp(-1/2)} \\ &\approx 0.205 \end{align}

Best Answer

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[Math] Actuarial Exam P problem (expected value of a max function.)

[Math] Finding probability that a car experiences a failure

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