Statistics: Multivariate Dist/Functions of RVs

[icestone111]

Homework Statement

1) A point (X1,X2,X3) are is chosen at random. X_1-3 are uniform distributions across the interval [0,1]
Determine P[(X₁-.5)² + X₂-.5)² + X₃-.5)² ≤ .25]

2) X and Y are random variables with the joint pdf: f(x,y) = 2(x + y) for 0≤x≤y≤1, 0 otherwise.
Find the pdf of Z = X + Y

2. The attempt at a solution

1) So I know the pdf for each one of the random variables is f(x) = 1 when 0≤x≤1
Imagining it in a grid, it would be the volume of the sphere with radius .5 centered at (.5,.5,.5). I am uncertain where to begin applying the pdf of each of the random variables to the graph of the probability needed to be found.

2) I know this has appeared once in another topic, but I can't figure out the bounds of this problem (if I'm not messing up elsewhere).
So the graph envisioned would be the points that lie above the line y=x, bounded by y <= 1, x >=0.
So what I would do from this point is attempt to find the solution for (1-x to 1)∫f(x,z-x)dx = (1-x to 1)∫2zdx = 2z -2z + 2zx = 2zx. What is wrong with this reasoning/where do I approach this problem after this?

 

[mighty2000]

1) To find the probability, we need to find the volume of the sphere with radius 0.5 centered at (0.5, 0.5, 0.5) within the given interval [0,1]. This can be done by integrating the pdf for each random variable over the given interval and then using the formula for the volume of a sphere. So the probability would be given by P[(X1-0.5)^2 + (X2-0.5)^2 + (X3-0.5)^2 ≤ 0.25] = ∫∫∫ f(x1)f(x2)f(x3) dx1dx2dx3 = (0 to 1)∫(0 to 1)∫(0 to 1) 1 dx1dx2dx3 = (0 to 1)∫(0 to 1) 1 dx2dx3 = (0 to 1) 1 dx3 = 1/8. So the probability is 1/8.

2) To find the pdf of Z, we first need to find the cumulative distribution function (CDF) of Z, which is given by Fz(z) = P(Z ≤ z) = P(X + Y ≤ z). Since X and Y are independent, we can write this as Fz(z) = P(X ≤ z - Y) = ∫∫ f(x,y) dx dy = (0 to z)∫(0 to z-x) 2(x+y) dy dx = (0 to z)∫ 2xz + 2x(z-x) dx = 2z^2 - z^3. Therefore, the pdf of Z is given by fz(z) = d/dz Fz(z) = 4z - 3z^2. So the pdf of Z is fz(z) = 4z - 3z^2 for 0 ≤ z ≤ 1.