Analyzing Dice Throw Statements

We are considering two independent random variables, $X$ and $Y$, representing the outcomes of throwing two fair dice. Each variable can take integer values from 1 to 6 with equal probability ($1/6$). The total number of possible outcomes when throwing two dice is $6 \times 6 = 36$. We need to evaluate the truthfulness of the given statements.

Evaluating Statement A: Conditional Probabilities

Statement A compares two conditional probabilities:

• Left Side: $P(X - Y = 0 | X + Y = 2)$
  • The condition $X + Y = 2$ implies the only possible outcome is $(X, Y) = (1, 1)$.
  • For this outcome, $X - Y = 1 - 1 = 0$.
  • Therefore, the probability $P(X - Y = 0 | X + Y = 2)$ is 1.
• Right Side: $P(X - Y = 0 | X + Y = 12)$
  • The condition $X + Y = 12$ implies the only possible outcome is $(X, Y) = (6, 6)$.
  • For this outcome, $X - Y = 6 - 6 = 0$.
  • Therefore, the probability $P(X - Y = 0 | X + Y = 12)$ is 1.

Since $1 = 1$, Statement A is TRUE.

Evaluating Statement B: Expectation of a Ratio

Statement B claims the expectation of the ratio $\frac{X-Y}{X+Y}$ is 0:

We can use the symmetry between $X$ and $Y$. For any outcome $(i, j)$ where $i \neq j$, the outcome $(j, i)$ is equally likely. The value of the expression for $(i, j)$ is $\frac{i-j}{i+j}$, and for $(j, i)$ it is $\frac{j-i}{j+i} = -\frac{i-j}{i+j}$. These values cancel each other out.

For outcomes where $i = j$, we have $X - Y = 0$, so the expression $\frac{X-Y}{X+Y} = \frac{0}{2i} = 0$.

Since all non-zero terms cancel in pairs and the terms where $X=Y$ are zero, the overall expectation is 0.

Therefore, Statement B is TRUE.

Evaluating Statement C: Covariance

Statement C relates the covariance of $(X+Y)$ and $(X-Y)$ to zero:

We use the property $Cov(U, V) = E(UV) - E(U)E(V)$.

• Let $U = X + Y$ and $V = X - Y$.
• $E(U) = E(X) + E(Y)$. Since $X$ and $Y$ are outcomes of fair dice, $E(X) = E(Y) = \frac{1+2+3+4+5+6}{6} = 3.5$. So, $E(U) = 3.5 + 3.5 = 7$.
• $E(V) = E(X) - E(Y) = 3.5 - 3.5 = 0$.
• $E(UV) = E((X + Y)(X - Y)) = E(X^2 - Y^2)$.
• $E(X^2) = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$.
• Since $X$ and $Y$ are identically distributed, $E(Y^2) = E(X^2) = \frac{91}{6}$.
• Therefore, $E(UV) = E(X^2) - E(Y^2) = \frac{91}{6} - \frac{91}{6} = 0$.
• $Cov(U, V) = E(UV) - E(U)E(V) = 0 - (7)(0) = 0$.

Alternatively, using covariance properties: $Cov(X+Y, X-Y) = Cov(X,X) - Cov(X,Y) + Cov(Y,X) - Cov(Y,Y) = Var(X) - 0 + 0 - Var(Y)$. Since $Var(X) = Var(Y)$, the covariance is 0.

Therefore, Statement C is TRUE.

Evaluating Statement D: Independence

Statement D claims $(X + Y)$ and $(X - Y)$ are independent.

Independence requires $P(A \cap B) = P(A)P(B)$ for all events A and B related to the variables.

Let's test this with specific events:

• Event A: $X+Y=7$. Outcomes are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$. $P(X+Y=7) = 6/36 = 1/6$.
• Event B: $X-Y=0$. Outcomes are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. $P(X-Y=0) = 6/36 = 1/6$.
• If independent, $P(X+Y=7 \text{ and } X-Y=0) = P(X+Y=7)P(X-Y=0) = (1/6)(1/6) = 1/36$.
• However, the condition $X+Y=7$ and $X-Y=0$ implies $2X=7$, so $X=3.5$. Since $X$ must be an integer, there are no outcomes satisfying both conditions. Thus, $P(X+Y=7 \text{ and } X-Y=0) = 0$.

Since $0 \neq 1/36$, the variables $(X+Y)$ and $(X-Y)$ are not independent.

Therefore, Statement D is FALSE.

Conclusion

Based on the analysis, statements A, B, and C are true.