For a certain product I pay $A$ amount of money. The lifetime of this product is a random variable $T$ from exponential distribution of parameter $\lambda>0.$ The manufacturer's policy is to replace ($100\%$ ) the product if it stops working in less that $a$ years. What is a logical amount of money I have to pay annually to an insurance company, for the time period that exceeds the $a$ years the manufacturer replaces the product?
Attempt. If $x$ is the desired amount, then for the time period $[a,T-a]$ I pay amount $x(T-a)$, as long as $T>a$, that is $x(T-a)1_{T>a}$. A fair deal would be something like: $$A=\mathbb{E}(x(T-a)1_{T>a})=x\int_{a}^{\infty}(t-a)\lambda e^{-\lambda t}\mathrm{d}t=x\,\frac{e^{-aλ}}{λ},$$ that is $x=Aλe^{aλ}.$
Is the above argumentation correct?
Thank for the help!
When you buy the product, you tend to insure against the loss which is defined as
Loss = $$L = 0 \text{ for} X<a$$
$$L = A \text{ for} X>a$$, for a moment assume it is A , as it should be less than that for the usage of the product up until "a". X is the time until first failure which is expoentially distributed.
Now you find the expected value of L over the life of the product and this is the amount you would need to buy insurance for.
$E(L) = 0.P(X<a) + A.P(X>a) = A P(X>a)$
$P(X>a) = 1 - \int_{0}^{a} \lambda e^{-\lambda x} dx = e^{-\lambda a}$
Thus the amount of capital that you have to buy protection for is $A_x = Ae^{-\lambda a}$. Equate this amount for an annuity $\hat a$ for selected number of $\hat x$ years. Thus the premium is