Consider two components whose lifetimes X1 and X2 are...

Consider two components whose lifetimes $X_{1}$ and $X_{2}$ are independent and exponentially distributed with parameters $\lambda_{1}$ and $\lambda_{2},$ respectively. Obtain the joint pdf of total lifetime $X_{1}+X_{2}$ and the proportion of total lifetime $X_{1} /\left(X_{1}+X_{2}\right)$ during which the first component operates.

Consider two components whose lifetimes $X_{1}$ and $X_{2}$ are independent and exponentially
distributed with parameters $\lambda_{1}$ and $\lambda_{2},$ respectively. Obtain the joint pdf of total lifetime $X_{1}+X_{2}$ and the proportion of total lifetime $X_{1} /\left(X_{1}+X_{2}\right)$ during which the first component operates.

Key Concepts

Exponential Distribution

This concept describes a continuous probability distribution characterized by a constant hazard rate and the memoryless property. An exponential random variable is often used to model the time until an event occurs, such as the lifetime of a component. Its probability density function is defined by a single parameter, typically denoted as ?, which signifies the rate at which events occur.

Independent Random Variables

When random variables are independent, the occurrence of one does not affect the probability distribution of the other. In the context of lifetime models, the independence assumption allows for the joint density to be expressed as the product of the individual densities, facilitating the analysis and transformation of combined variables.

Joint Probability Density Function (pdf)

The joint pdf is a function that provides the likelihood of two continuous random variables occurring simultaneously within a specific range. In problems involving multiple variables, such as lifetimes of different components, establishing the joint pdf is crucial for deriving properties of functions of those variables, such as their sum and ratios.

Transformation of Variables

Transformation techniques are used to change from one set of random variables to another, particularly when the new variables represent meaningful summaries such as the total lifetime or the ratio of lifetimes. This requires expressing the original variables in terms of the new variables and computing the resulting pdf through methods such as the change of variables formula.

Jacobian Determinant

In the process of transforming random variables, the Jacobian determinant is used to adjust the scale of the probability density function appropriately. It accounts for the local stretching or shrinking of the space under the transformation, ensuring that the total probability is preserved in the new variable space.

Transcript

00:01 So for this problem, we want to start out by finding the joint pdf of x1 plus x2.

00:06 So we're using parameters and we have a pdf of x1 and a pdf of x2.

00:22 And both of the pdfs are the same.

00:25 So i'm going to write it like this, but you could do two separate ones.

00:29 And so we have parameter x because we're denoted this from one and two so if this was for two we would just say parameter two so we have parameter x times e times negative parameter lambda lambda y x1 and this is greater than some aspect of x so if this was x1 or x2 and this and there it's greater than zero and when we say that we're saying that it had to exist or it's otherwise zero and when you're trying to find a new random variable you're doing because that's what we're going to be doing you do you equals x1 plus x2 and then for v you have x1 over x1 plus x2 so let's keep that and let's remember that so we want to find the transformation and we're going to use this right here and so first we're going to consider v and then you so or we can start with you you is equal to x1 plus x2 so x2 is you and we're solving for you right now i mean we're solving for x squared so x squared it's u minus uv and the reason why that is is because you're going to see why in a second of why x1 changes i just wanted to show you kind of what we're going to be doing how we're going to be solving and so when we get to v it's x1 plus and you can't know v without you but i just want to show you the pattern so it's x1 plus x2 so when we're solving for x1 we bring it to that side and we have x1 plus x2 because we brought this one down there so next we are just going to solve.

03:11 And so x1 plus x2 is what? boom.

03:21 So this is supposed to be times.

03:26 So we have x1 is equal to vu.

03:33 And that's uv.

03:35 And so that's how when we get x1, then that becomes that.

03:38 And so that's how we find those random variables.

03:42 And so when we have that, we have let's write our new little piece of information that we need x1 is equal to uv and x2 is equal to uv u minus uv so now we need to calculate the jocobian transition and it's really hard and complicated well it's not hard and complicated but pretty much we have v here.

04:24 We have u...

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