Question 1163974: The random variables (X,Y) have joint pdf f(x,y) = 1/32 for the region 0 < x < 8, 0 < y < 4. The pdf is zero outside this region.
i. What are the marginal distributions of X and Y?
ii. Find the joint cdf of (X,Y) F(x,y).
iii. Find the probability P(X+Y < 4).
iv. Find the conditional probability density f(x|y)
v. Find the mean of X given Y i.e. E(X|Y).
Answer by CPhill(2264)   (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=1
# Let's perform some basic checks for the probability and expectations
# f(x,y) = 1/32 for 0
# iii. P(X+Y < 4)
# Triangle with vertices (0,0), (4,0), (0,4)
area_event = 0.5 * 4 * 4 # 8
prob_event = area_event * (1/32) # 8/32 = 1/4
# v. E(X|Y)
# f(x|y) = f(x,y)/f_Y(y)
# f_Y(y) = integral_0^8 (1/32) dx = 8/32 = 1/4 for 0
# f(x|y) = (1/32) / (1/4) = 4/32 = 1/8 for 0
# E(X|Y) = integral_0^8 x * (1/8) dx = [x^2/16]_0^8 = 64/16 = 4.
print(f"P(X+Y < 4): {prob_event}")
print(f"E(X|Y): {4.0}")
```
```text?code_stdout&code_event_index=1
P(X+Y < 4): 0.25
E(X|Y): 4.0
```
This is a uniform distribution over a rectangular region. Let's solve each part step-by-step.
### **i. What are the marginal distributions of $X$ and $Y$?**
To find the marginals, we integrate the joint pdf over the range of the other variable.
* **For $X$ ($0 < x < 8$):**
$$f_X(x) = \int_0^4 \frac{1}{32} \, dy = \left[ \frac{y}{32} \right]_0^4 = \frac{4}{32} = \frac{1}{8}$$
* **For $Y$ ($0 < y < 4$):**
$$f_Y(y) = \int_0^8 \frac{1}{32} \, dx = \left[ \frac{x}{32} \right]_0^8 = \frac{8}{32} = \frac{1}{4}$$
Since the marginals are constant, **$X \sim \text{Uniform}(0, 8)$** and **$Y \sim \text{Uniform}(0, 4)$**.
---
### **ii. Find the joint CDF $F(x,y)$**
The joint CDF is the integral of the joint pdf from the lower bounds to $(x, y)$ within the region:
$$F(x, y) = \int_0^y \int_0^x \frac{1}{32} \, du \, dv = \frac{xy}{32}$$
This is valid for $0 < x < 8$ and $0 < y < 4$. (For values outside this range, the CDF is $0$ if $x,y \le 0$, and $1$ if $x \ge 8, y \ge 4$).
---
### **iii. Find the probability $P(X+Y < 4)$**
Since the distribution is uniform, the probability is simply the **area of the event region** divided by the **total area** ($8 \times 4 = 32$).
* The region $X+Y < 4$ forms a triangle with vertices at $(0,0)$, $(4,0)$, and $(0,4)$.
* $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8$.
* **$P(X+Y < 4) = \frac{8}{32} = \frac{1}{4}$ (or $0.25$)**.
---
### **iv. Find the conditional probability density $f(x \mid y)$**
The conditional density is defined as the joint density divided by the marginal density of the conditioning variable:
$$f(x \mid y) = \frac{f(x, y)}{f_Y(y)} = \frac{1/32}{1/4} = \frac{4}{32} = \frac{1}{8}$$
This is valid for $0 < x < 8$. Note that $f(x \mid y)$ does not depend on $y$, which proves that **$X$ and $Y$ are independent**.
---
### **v. Find the mean of $X$ given $Y$ ($E[X \mid Y]$)**
Since we found $f(x \mid y) = 1/8$ (a uniform distribution on $[0, 8]$):
$$E[X \mid Y] = \int_0^8 x \cdot f(x \mid y) \, dx = \int_0^8 \frac{x}{8} \, dx$$
$$E[X \mid Y] = \left[ \frac{x^2}{16} \right]_0^8 = \frac{64}{16} = 4$$
Because $X$ and $Y$ are independent, $E[X \mid Y]$ is simply the same as the marginal mean $E[X]$, which is the midpoint of its range $(0+8)/2 = 4$.
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