Using the Joint MGF to calculate a probability

Question

This may seem like a dumb question, but I don't know how to approach it well at all.

A machine consisting of two components will fail if one of the components fails.

Let $X$ be the time-to-failure random variable of the first component and $Y$ be the time-to-failure random variable of the second component. The time-to-failure is measured in hours. The time-to-failure of the two components have the following joint moment generating function:

$$ M_{X,Y}(s,t)=\frac{1}{(1-3s-2t+6ts)} $$

Calculate the probability that the machine will fail within the first hour.

Answer 1

We know that the mgf corresponds to only one probability function. Therefore we want to find the form of the mgf in question. Looking up in tables or by memory, we identify a form similar to the one of an exponential mgf:

$$ M_x(t)=\frac{1}{1-\theta t} $$

For an exponential of the form

$$ \frac{1}{\theta}e^{-(1/\theta )x} $$

Also we can assume that the behavior of the components is independent from each other, so the product of the moment generating functions is the joint moment generating function. Solving a simple equation:

$$ \frac{1}{(1-\theta_1 t)}\frac{1}{(1-\theta_2 t)}=\frac{1}{1-3s-2t-6ts} $$

Giving that $\theta_1=2$ and $\theta_2 = 3$. Therefore we want to calculate the probability of either component $X$ or $Y$ lasting less than some time $k$. Therefore we want to consider the minimum of either of the distributions failing before time $k$. We also know that $\min\{X_i\}$ when those $X_i$ are exponentials, the new minimum random variable has a distribution of the products, which should be easy to follow from here.

Just in case, The distribution of the minimum of the exponentials is an exponential with the parameter $\lambda_{min}=\sum \lambda_i$, where $\lambda_i=\frac{1}{\theta_i}$ because we used the alternative parametrization of the exponential in the problem.

Answer 2

A start: Note that the joint moment generating function is $$\frac{1}{3s-1}\cdot \frac{1}{2t-1}.$$

Recall that an exponentially distributed random variable with parameter $\lambda$ has mgf $\frac{\lambda}{x-\lambda}$, that is, $\frac{1}{x/\lambda-1}$. We want the probability that the minimum of the lives is less than an hour. It is easier to find first the probability both are alive after an hour.