To find the probability density function (pdf) of $Z = X + Y$ when $X$ and $Y$ are independent random variables with a given pdf $f(x) = 2x$ for $0 < x < 1$, follow these steps:

---

Step 1: Identify the range of $Z$

The range of $Z = X + Y$ is determined by the minimum and maximum values of $X + Y$:

• Since $0 < X, Y < 1$, the sum $Z = X + Y$ ranges from $0$ to $2$.

---

Step 2: Derive the pdf of $Z$

The pdf of $Z$, $f_Z(z)$, is obtained using the convolution formula for independent random variables: $f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z - x) \, dx.$ Substitute $f_X(x) = 2x$ and $f_Y(y) = 2y$, valid for $0 < x, y < 1$, and compute $f_Z(z)$ piecewise depending on $z$:

Case 1: $0 < z \leq 1$

Here, $0 < x < z$, so: $f_Z(z) = \int_0^z 2x \cdot 2(z - x) \, dx = 4 \int_0^z (xz - x^2) \, dx.$ Expand the integral: $f_Z(z) = 4 \left[ \frac{z x^2}{2} - \frac{x^3}{3} \right]_0^z = 4 \left( \frac{z^3}{2} - \frac{z^3}{3} \right).$ Simplify: $f_Z(z) = 4 \cdot \frac{z^3}{6} = \frac{2z^3}{3}.$

Case 2: $1 < z \leq 2$

Here, $z - 1 < x < 1$, so: $f_Z(z) = \int_{z-1}^1 2x \cdot 2(z - x) \, dx = 4 \int_{z-1}^1 (xz - x^2) \, dx.$ Expand the integral: $f_Z(z) = 4 \left[ \frac{z x^2}{2} - \frac{x^3}{3} \right]_{z-1}^1.$ Evaluate at the limits:

• At $x = 1$: $\frac{z (1)^2}{2} - \frac{(1)^3}{3} = \frac{z}{2} - \frac{1}{3}$.
• At $x = z-1$: $\frac{z (z-1)^2}{2} - \frac{(z-1)^3}{3}$.

Substitute and simplify: $f_Z(z) = 4 \left[ \left( \frac{z}{2} - \frac{1}{3} \right) - \left( \frac{z (z-1)^2}{2} - \frac{(z-1)^3}{3} \right) \right].$

Simplify Piecewise Formula

Combine the cases:

\begin{cases} \frac{2z^3}{3}, & 0 < z \leq 1, \\ \text{Piecewise computation}, & 1 < z \leq 2. \end{cases}$$ --- ### Step 3: Sketch the Graph The graph will involve a cubic increase in $$0 < z \leq 1$$ and a decay from $$1 < z \leq 2$$. --- Let me compute the exact formula for the second piece and plot the graph.import numpy as np import matplotlib.pyplot as plt # Define piecewise pdf for Z = X + Y def f_Z(z): if 0 < z <= 1: return (2/3) * z**3 elif 1 < z <= 2: return (4/3) * (z - 1)**3 - (4/3) * (z - 1)**2 * z + (2/3) * z - (2/3) else: return 0 # Generate z values for plotting z_values = np.linspace(0, 2, 500) pdf_values = [f_Z(z) for z in z_values] # Plot the pdf plt.figure(figsize=(8, 5)) plt.plot(z_values, pdf_values, label="PDF of Z = X + Y", linewidth=2) plt.title("PDF of Z = X + Y") plt.xlabel("z") plt.ylabel("f_Z(z)") plt.grid(True) plt.legend() plt.show() <<ImageDisplayed>>successThe graph above shows the pdf of $$Z = X + Y$$. It increases cubically from 0 to 1 and then decays nonlinearly between 1 and 2. Let me know if you'd like further details or specific steps explained. Here are 5 related questions to expand on this topic: 1. How would the result change if $$X$$ and $$Y$$ had different distributions? 2. Can you generalize this process for finding the sum of two random variables with arbitrary distributions? 3. How do you verify that the resulting pdf integrates to 1? 4. What is the expected value of $$Z$$ in this case? 5. How does the shape of the pdf change if $$X$$ and $$Y$$ are dependent? **Tip:** Convolution is a powerful tool for combining independent random variables. Practice with different types of distributions to strengthen your understanding.