Hi
* Suppose X has the pdf f(x)=(1/2)e^-|x| if -∞<x<∞
a)find the pdf of Y=|X|^3
b) Let W=0 if x<=0 and W=1 if x>1. Find cdf (.dist.func) of W.
Hi!
Sure, I can help you with that.
a) To find the PDF of Y=|X|^3, we can start by finding the cumulative distribution function (CDF) of Y and then differentiating it to obtain the PDF.
1. First, let's find the CDF of Y:
The CDF of Y, denoted as F_Y(y), is defined as the probability that Y is less than or equal to y.
So, P(Y ≤ y) = P(|X|^3 ≤ y)
Since Y=|X|^3, we can rewrite this as:
P(|X| ≤ y^(1/3))
Now, let's split this into two cases:
Case 1: y ≥ 0
For this case, |X| ≤ y^(1/3) holds true for all values of X since |X| is always greater than or equal to 0. Therefore, P(|X| ≤ y^(1/3)) = 1.
Case 2: y < 0
For this case, |X| ≤ y^(1/3) does not hold true for any values of X since |X| is always greater than or equal to 0. Therefore, P(|X| ≤ y^(1/3)) = 0.
Putting these cases together, we can define the CDF of Y as:
F_Y(y) = 1 for y ≥ 0, and F_Y(y) = 0 for y < 0.
2. Now, let's differentiate the CDF of Y to find the PDF of Y:
The PDF, denoted as f_Y(y), is the derivative of the CDF with respect to y.
For y ≥ 0, the derivative of F_Y(y) is 0 since F_Y(y) is constant.
For y < 0, the derivative of F_Y(y) is also 0 since F_Y(y) is constant.
Therefore, the PDF of Y is:
f_Y(y) = 0 for y ≥ 0, and f_Y(y) = 0 for y < 0.
b) To find the CDF of W, we can use the definition of the CDF.
The CDF of W, denoted as F_W(w), is defined as the probability that W is less than or equal to w.
So, P(W ≤ w) = P(W = 0) + P(W = 1)
To find P(W = 0) and P(W = 1), we can consider the cases based on the value of x.
Case 1: x ≤ 0
For this case, P(W = 0) = 1 since W is defined to be 0 when x ≤ 0, and P(W = 1) = 0.
Case 2: x > 0
For this case, P(W = 0) = 0 since W is defined to be 1 when x > 0, and P(W = 1) = 1.
Putting these cases together, we can define the CDF of W as:
F_W(w) = 1 for w > 0, and F_W(w) = 0 for w ≤ 0.
I hope this helps! Let me know if you have any further questions.