Let U, V be random numbers chosen independently from the interval [0; 1] with uniform distribution. Find the cumulative distribution and density of each of the variables (a) Y = U + V. (b) Y = Absolute value of (U - V).
To find the cumulative distribution function (CDF) and probability density function (PDF) of the random variables Y = U + V and Y = |U - V|, we can use the properties of random variables and the transformation technique.
(a) Y = U + V:
To find the CDF of Y, we can calculate P(Y <= y), where y is any given number.
1. For y < 0:
Since U and V are both random numbers chosen independently from the interval [0, 1], the sum U + V will always be greater than or equal to zero. So, P(Y <= y) = 0.
2. For 0 <= y < 1:
To find P(Y <= y), we need to consider all possible combinations of U and V that give a sum less than or equal to y.
- If U < y, then V must be in the interval [0, y - U] for Y to be less than or equal to y.
- The probability of U < y is P(U < y) = y, since U is uniformly distributed in the interval [0, 1].
- The length of the interval [0, y - U] is y - U.
- Hence, the probability of V being in the interval [0, y - U] is P(V < y - U) = y - U.
By considering all possible values of U, the CDF expression becomes:
P(Y <= y) = ∫[0, y] (y - U) dU = y^2/2.
3. For y >= 1:
Since U and V are both random numbers chosen independently from [0, 1], the sum U + V can be greater than 1 if either U or V is greater than 1/2. The probability of either U or V being greater than 1/2 is P(U > 1/2 or V > 1/2) = 1 - P(U <= 1/2 and V <= 1/2) = 1 - (1/2 * 1/2) = 3/4
Therefore, the CDF of Y = U + V is given by:
F(y) = 0 for y < 0,
y^2/2 for 0 <= y < 1,
3/4 + (y - 1)^2/2 for y >= 1.
To find the PDF, we can differentiate the CDF with respect to y:
f(y) = d/dy F(y) = y for 0 <= y < 1,
y - 1 for y >= 1,
0 otherwise.
(b) Y = |U - V|:
To find the CDF and PDF of the absolute difference |U - V|, we can use similar techniques as above.
1. For y < 0:
Since the absolute value is always non-negative, P(Y <= y) = 0.
2. For 0 <= y < 1:
To find P(Y <= y), we need to consider all possible combinations of U and V that give an absolute difference less than or equal to y.
- If |U - V| <= y, then we have two cases:
a. U >= V:
- U - V <= y
- U <= y + V
- The probability of U <= y + V is P(U <= y + V) = (y + V).
- The length of the interval [0, V] is V.
- Hence, the probability of U being in the interval [0, V] is P(U <= V) = V.
b. U < V:
- V - U <= y
- U >= V - y
- The probability of U >= V - y is P(U >= V - y) = 1 - (V - y) = y + 1 - V.
- The length of the interval [V, 1] is 1 - V.
- Hence, the probability of U being in the interval [V, 1] is P(U >= V - y) = y + 1 - V.
By considering all possible values of V, the CDF expression becomes:
P(Y <= y) = ∫[0, y] (y + V) dV + ∫[y, 1] (y + 1 - V) dV = y^2 + y - y^2/2 = y + y^2/2.
3. For y >= 1:
Using similar reasoning as in part (a), P(Y <= y) = 3/4 + (y - 1)^2/2.
Hence, the CDF of Y = |U - V| is given by:
G(y) = 0 for y < 0,
y + y^2/2 for 0 <= y < 1,
3/4 + (y - 1)^2/2 for y >= 1.
To find the PDF, we can differentiate the CDF with respect to y:
g(y) = d/dy G(y) = 1 + y for 0 <= y < 1,
y - 1 for y >= 1,
0 otherwise.
Therefore, the PDF of Y = |U - V| is given by:
g(y) = 1 + y for 0 <= y < 1,
y - 1 for y >= 1,
0 otherwise.