Question 1172515: Suppose that the joint density function of the random variables X and Y is given by
F(x,y) ={8xy if 0≤y≤x≤1
0 elsewhere
a)ComputeP(X+Y <1).(Sketch the region clearly.)
b)Find the marginal density of X, i.e.fX(x) and marginal density of Y, i.e.fY(y).
c)Find the conditional density of Y given X=1/2, that is,fY|X(y|1/2).
d)Find the conditional expectation of Y given X=1/2, that is,E[Y|X=1/2].
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step by step.
**Given:**
* Joint density function:
* F(x, y) = 8xy if 0 ≤ y ≤ x ≤ 1
* F(x, y) = 0 elsewhere
**a) Compute P(X + Y < 1)**
1. **Sketch the Region:**
* The region defined by 0 ≤ y ≤ x ≤ 1 is a triangle in the first quadrant with vertices (0, 0), (1, 0), and (1, 1).
* The condition X + Y < 1 implies Y < 1 - X. This is the region below the line Y = 1 - X.
* We need to find the intersection of these two regions.
* The intersection is a triangle with vertices (0, 0), (1, 0), and (1/2, 1/2).
2. **Set up the Integral:**
* P(X + Y < 1) = ∫∫ F(x, y) dy dx
* The limits of integration are:
* 0 ≤ x ≤ 1/2
* 0 ≤ y ≤ 1 - x
3. **Evaluate the Integral:**
* P(X + Y < 1) = ∫(from 0 to 1/2) ∫(from 0 to 1-x) 8xy dy dx
* P(X + Y < 1) = ∫(from 0 to 1/2) [4xy^2](from 0 to 1-x) dx
* P(X + Y < 1) = ∫(from 0 to 1/2) 4x(1 - x)^2 dx
* P(X + Y < 1) = ∫(from 0 to 1/2) 4x(1 - 2x + x^2) dx
* P(X + Y < 1) = ∫(from 0 to 1/2) (4x - 8x^2 + 4x^3) dx
* P(X + Y < 1) = [2x^2 - (8/3)x^3 + x^4](from 0 to 1/2)
* P(X + Y < 1) = 2(1/4) - (8/3)(1/8) + (1/16)
* P(X + Y < 1) = 1/2 - 1/3 + 1/16
* P(X + Y < 1) = (24 - 16 + 3)/48 = 11/48
**b) Find the Marginal Densities fX(x) and fY(y)**
1. **Marginal Density of X (fX(x)):**
* fX(x) = ∫(from 0 to x) 8xy dy
* fX(x) = [4xy^2](from 0 to x)
* fX(x) = 4x^3, 0 ≤ x ≤ 1
2. **Marginal Density of Y (fY(y)):**
* fY(y) = ∫(from y to 1) 8xy dx
* fY(y) = [4x^2y](from y to 1)
* fY(y) = 4y - 4y^3, 0 ≤ y ≤ 1
**c) Find the Conditional Density fY|X(y|1/2)**
1. **Conditional Density Formula:**
* fY|X(y|x) = F(x, y) / fX(x)
2. **Substitute x = 1/2:**
* fY|X(y|1/2) = 8(1/2)y / 4(1/2)^3
* fY|X(y|1/2) = 4y/(1/2)
* fY|X(y|1/2) = 8y.
* Since 0 <= y <= x, and x = 1/2, then 0<=y<=1/2.
3. **Result:**
* fY|X(y|1/2) = 8y, 0 ≤ y ≤ 1/2
**d) Find the Conditional Expectation E[Y|X=1/2]**
1. **Conditional Expectation Formula:**
* E[Y|X=1/2] = ∫ y * fY|X(y|1/2) dy
2. **Substitute and Evaluate:**
* E[Y|X=1/2] = ∫(from 0 to 1/2) y * 8y dy
* E[Y|X=1/2] = ∫(from 0 to 1/2) 8y^2 dy
* E[Y|X=1/2] = [(8/3)y^3](from 0 to 1/2)
* E[Y|X=1/2] = (8/3) * (1/8)
* E[Y|X=1/2] = 1/3
**Summary**
* a) P(X + Y < 1) = 11/48
* b) fX(x) = 4x^3, 0 ≤ x ≤ 1; fY(y) = 4y - 4y^3, 0 ≤ y ≤ 1
* c) fY|X(y|1/2) = 8y, 0 ≤ y ≤ 1/2
* d) E[Y|X=1/2] = 1/3
Answer by ikleyn(52903)  (Show Source):
You can put this solution on YOUR website! .
Suppose that the joint density function of the random variables X and Y is given by
F(x,y) ={8xy if 0≤y≤x≤1
0 elsewhere
a)ComputeP(X+Y <1).(Sketch the region clearly.)
b)Find the marginal density of X, i.e.fX(x) and marginal density of Y, i.e.fY(y).
c)Find the conditional density of Y given X=1/2, that is,fY|X(y|1/2).
d)Find the conditional expectation of Y given X=1/2, that is,E[Y|X=1/2].
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In the post by @CPhill, the region for integration is determined INCORRECTLY,
therefore, all his subsequent calculations are INCORRECT and IRRELEVANT.
Indeed, at the end of n.1, @CPhill writes
* The intersection is a triangle with vertices (0, 0), (1, 0), and (1/2, 1/2).
It is INCORRECT. Actually, this triangle has vertices (0,0), (1,0) and (0,1).
As I said, everything that follows in the post by @CPhill is WRONG.
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Regarding the post by @CPhill . . .
Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.
The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.
It has no feeling of shame - it is shameless.
This time, again, it made an error.
Although the @CPhill' solution are copy-paste Google AI solutions, there is one essential difference.
Every time, Google AI makes a note at the end of its solutions that Google AI is experimental
and can make errors/mistakes.
All @CPhill' solutions are copy-paste of Google AI solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, he NEVER SAYS TRUTH.
Every time, @CPhill embarrassed to tell the truth.
But I am not embarrassing to tell the truth, as it is my duty at this forum.
And the last my comment.
When you obtain such posts from @CPhill, remember, that NOBODY is responsible for their correctness,
until the specialists and experts will check and confirm their correctness.
Without it, their reliability is ZERO and their creadability is ZERO, too.
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