To solve the problem of finding the probability that $Z = X + Y \leq -2$, where $X$ and $Y$ are statistically independent random variables uniformly distributed in the ranges $(-1, 1)$ and $(-2, 1)$ respectively, let's go step by step.

1. Definitions and Ranges

• $X$ is uniformly distributed over $(-1, 1)$. $f_X(x) = \frac{1}{2}, \quad -1 < x < 1$
• $Y$ is uniformly distributed over $(-2, 1)$. $f_Y(y) = \frac{1}{3}, \quad -2 < y < 1$

2. Joint Distribution

Since $X$ and $Y$ are independent, the joint probability density function (PDF) of $X$ and $Y$ is: $f_{X,Y}(x, y) = f_X(x) \cdot f_Y(y) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}, \quad -1 < x < 1, \ -2 < y < 1$

3. Probability Calculation

We want to find the probability $P(Z \leq -2)$, where $Z = X + Y$.

Let's consider the inequality: $Z \leq -2 \quad \text{which implies} \quad X + Y \leq -2$

Given the ranges of $X$ and $Y$:

• $X$ ranges from $-1$ to $1$.
• $Y$ ranges from $-2$ to $1$.

For $Z = X + Y \leq -2$, we can express $Y$ in terms of $X$: $Y \leq -2 - X$

The probability can be computed by integrating the joint PDF over the region defined by the above inequality:

4. Integration Boundaries

1. For $X = -1$: $Y \leq -2 - (-1) = -1$ Since $Y$ must be between $-2$ and $1$, this condition is valid.

2. For $X = 1$: $Y \leq -2 - 1 = -3$ Since $Y$ is bounded between $-2$ and $1$, this condition is never satisfied.

The region of integration is: $\int_{-1}^{0} \int_{-2}^{-2-x} \frac{1}{6} \, dy \, dx$

5. Perform the Integration

First, integrate with respect to $Y$: $\int_{-2}^{-2-x} \frac{1}{6} \, dy = \frac{1}{6} \left[ y \right]_{-2}^{-2-x} = \frac{1}{6}(-2-x + 2) = \frac{-x}{6}$

Now, integrate with respect to $X$: $\int_{-1}^{0} \frac{-x}{6} \, dx = \frac{-1}{6} \int_{-1}^{0} x \, dx = \frac{-1}{6} \left[ \frac{x^2}{2} \right]_{-1}^{0}$ $= \frac{-1}{6} \left( 0 - \frac{1}{2} \right) = \frac{1}{12}$

6. Final Probability

Thus, the probability that $Z \leq -2$ is: $P(Z \leq -2) = \frac{1}{12}$

Conclusion

The probability that $Z \leq -2$ is $\frac{1}{12}$.

Would you like any more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the probability that $Z$ equals a specific value, say $Z = 0$?
2. How does the range of $Z = X + Y$ change if the distribution of $X$ or $Y$ changes?
3. What is the expected value of $Z = X + Y$?
4. How do you compute the probability $Z$ is within a certain range, say $-1 \leq Z \leq 1$?
5. What is the variance of $Z = X + Y$?

Tip: For independent random variables, the sum of their means gives the mean of their sum, and the sum of their variances gives the variance of their sum.