What Is the Probability Density Function of Z=|X-y| in a Uniformly Broken Stick?

[de1337ed]

Homework Statement

Consider a stick with unit length. We break at a point, the distance from which to the left end is a random variable X.

Find the PDF of Z=|X-y|, where y is a set point between [0,1] (hint: define an event A and write f_Z(z) = P(A)f_Z|A(z) + P(A^c)f_Z|A^c(z)).

The Attempt at a Solution

So basically, I set A to be the event {X<y}, so
P(A) = ∫1dx from 0 to y = y.
P(A^c) = ∫1dx from y to 1 = 1-y.

Now, I need to find f_Z|A(z) and f_Z|A^c(z)
I think my fundamentals are bad. Can someone please explain to me in words what the PDF is? Is it the probability of getting getting a value in a range of values? What exactly does the PDF mean in this problem?
Can someone try and talk to me through finding f_Z|A(z)?
Thank you.

 

[mighty2000]

Hello there! It seems like you are on the right track with your approach. The PDF, or probability density function, is a mathematical function that describes the probability distribution of a continuous random variable. In simpler terms, it tells you the likelihood of getting a certain value for a variable in a given range.

In this problem, we are trying to find the PDF of Z, which is the absolute value of the difference between X and y. To do this, we can break it down into two cases - when X is less than y (event A) and when X is greater than y (event Ac).

For event A, we can think of it as the stick breaking at a point that is closer to the left end than the set point y. Since X is a random variable with a uniform distribution, the probability of this happening is simply the length of the segment from 0 to y, which is y. So, P(A) = y.

Now, to find fZ|A(z), we need to find the probability of Z taking on a specific value z, given that event A has occurred. In other words, we want to find the probability that the difference between X and y is equal to z, given that X is less than y. This can be written as P(Z=z|A).

To find this probability, we can use the fact that X has a uniform distribution on the interval [0,y]. This means that the probability of X taking on a specific value within this interval is the same for all values. So, we can say that P(Z=z|A) = P(X=y-z|A) = 1/y. This is because for Z to equal z, X must be equal to y-z (since we are taking the absolute value). And since X has a uniform distribution on the interval [0,y], the probability of X taking on a specific value within this interval is 1/y.

So, we can write fZ|A(z) = P(Z=z|A) = 1/y for z between 0 and y (since Z can only take on values between 0 and y for event A to occur).

Similarly, for event Ac, we can think of it as the stick breaking at a point that is further away from the left end than the set point y. In this case, the probability of this happening is simply the length of the segment from y to 1, which is