Suppose that the p.d.f. of a random variable X is given by f(x) = (1/2)x for 0 < x < 2, and f(x) = 0 otherwise. Also, suppose that Y = X(2 - X). Determine the CDF and the PDF of Y.
Solution
To solve this problem without using the Jacobian, we will first determine the cumulative distribution function (CDF) and probability density function (PDF) of the transformed random variable Y=X(2−X) given the PDF of X.
Given:
The PDF of X is:
\frac{1}{2}x & \text{for } 0 < x < 2, \\
0 & \text{otherwise}.
\end{cases}$$
- We need to find the CDF and PDF of $$Y = X(2 - X)$$.
#### Step 1: Range of $$Y$$
We first determine the range of $$Y$$ by analyzing the transformation $$Y = X(2 - X)$$.
This function is quadratic in $$X$$, and can be rewritten as:
$$Y = 2X - X^2.$$
To find the range of $$Y$$, we determine the maximum and minimum values of $$Y$$ over the domain $$0 < x < 2$$.
- The derivative of $$Y$$ with respect to $$X$$ is:
$$\frac{dY}{dX} = 2 - 2X.$$
- Setting the derivative to 0 to find the critical points:
$$2 - 2X = 0 \implies X = 1.$$
At $$X = 1$$, the value of $$Y$$ is:
$$Y(1) = 1(2 - 1) = 1.$$
Now we compute $$Y$$ at the endpoints of the interval $$0 < x < 2$$:
- For $$X = 0$$, we get:
$$Y(0) = 0(2 - 0) = 0.$$
- For $$X = 2$$, we get:
$$Y(2) = 2(2 - 2) = 0.$$
So the range of $$Y$$ is $$0 \leq Y \leq 1$$.
#### Step 2: Finding the CDF of $$Y$$
The CDF of $$Y$$, denoted as $$F_Y(y)$$, is given by:
$$F_Y(y) = P(Y \leq y) = P(X(2 - X) \leq y).$$
This requires solving for $$X$$ in terms of $$Y$$. The quadratic equation $$Y = X(2 - X)$$ simplifies to:
$$X^2 - 2X + Y = 0.$$
Using the quadratic formula:
$$X = \frac{2 \pm \sqrt{4 - 4Y}}{2} = 1 \pm \sqrt{1 - Y}.$$
Since $$0 < X < 2$$, we take the solution $$X = 1 - \sqrt{1 - Y}$$. So,
$$F_Y(y) = P\left( X \leq 1 - \sqrt{1 - y} \right).$$
We now find the probability for $$X$$ in the interval $$(0, 2)$$. Using the PDF of $$X$$, the CDF of $$Y$$ becomes:
$$F_Y(y) = \int_0^{1 - \sqrt{1 - y}} f_X(x) \, dx = \int_0^{1 - \sqrt{1 - y}} \frac{1}{2}x \, dx.$$
This is the integral of $$\frac{1}{2}x$$, which evaluates as follows:
$$F_Y(y) = \left[ \frac{1}{4}x^2 \right]_0^{1 - \sqrt{1 - y}} = \frac{1}{4}(1 - \sqrt{1 - y})^2.$$
#### Step 3: Finding the PDF of $$Y$$
The PDF $$f_Y(y)$$ is the derivative of the CDF $$F_Y(y)$$ with respect to $$y$$:
$$f_Y(y) = \frac{d}{dy} \left( \frac{1}{4}(1 - \sqrt{1 - y})^2 \right).$$
Using the chain rule to differentiate, we get:
$$f_Y(y) = \frac{1}{4} \cdot 2(1 - \sqrt{1 - y}) \cdot \left( \frac{1}{2}(1 - y)^{-\frac{1}{2}} \right).$$
Simplifying this expression:
$$f_Y(y) = \frac{1 - \sqrt{1 - y}}{2\sqrt{1 - y}}.$$
Thus, the PDF of $$Y$$ is:
$$f_Y(y) = \frac{1 - \sqrt{1 - y}}{2\sqrt{1 - y}} \quad \text{for } 0 < y < 1.$$
### Summary:
- The CDF of $$Y$$ is:
$$F_Y(y) = \frac{1}{4}(1 - \sqrt{1 - y})^2.$$
- The PDF of $$Y$$ is:
$$f_Y(y) = \frac{1 - \sqrt{1 - y}}{2\sqrt{1 - y}} \quad \text{for } 0 < y < 1.$$
Let me know if you'd like further clarifications or details on any step! Here are some follow-up questions for deeper understanding:
1. What is the significance of the quadratic transformation of $$X$$ to $$Y$$?
2. How does the support of $$X$$ influence the support of $$Y$$?
3. Why did we choose the solution $$X = 1 - \sqrt{1 - y}$$ and not $$X = 1 + \sqrt{1 - y}$$?
4. How would the PDF change if the range of $$X$$ were different?
5. Can you explain the importance of the CDF when finding the PDF of a transformed variable?
**Tip:** When solving for the distribution of a transformed variable, it is often helpful to first understand the range and behavior of the transformation before diving into calculations.
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Math Problem Analysis
Mathematical Concepts
Probability Cumulative Distribution Function Probability Density Function Transformation of Variables
Formulas
f_X(x) = (1/2)x for 0 < x < 2 Transformation: Y = X(2 - X) Quadratic formula: ax^2 + bx + c = 0 CDF: F_Y(y) = P(Y ≤ y) PDF: f_Y(y) = d/dy[F_Y(y)]
Theorems
Transformation of random variables Quadratic transformation in probability