1. **Problem Statement:** Calculate the covariance $\mathrm{Cov}(X,Y)$ given the joint probability distribution table: | X \ Y | 0 | 1 | 2 | |-------|---|---|---| | 0 | \frac{1}{9} | \frac{2}{9} | \frac{1}{9} | | 1 | \frac{2}{9} | \frac{2}{9} | 0 | | 2 | \frac{1}{9} | 0 | 0 | 2. **Formula:** $$\mathrm{Cov}(X,Y) = E[XY] - E[X]E[Y]$$ where $E[XY]$ is the expected value of the product $XY$, and $E[X]$, $E[Y]$ are the expected values of $X$ and $Y$ respectively. 3. **Calculate $E[X]$:** $$E[X] = \sum_x x P_X(x)$$ First find marginal $P_X(x)$ by summing over $Y$: - $P_X(0) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9}$ - $P_X(1) = \frac{2}{9} + \frac{2}{9} + 0 = \frac{4}{9}$ - $P_X(2) = \frac{1}{9} + 0 + 0 = \frac{1}{9}$ Then: $$E[X] = 0 \times \frac{4}{9} + 1 \times \frac{4}{9} + 2 \times \frac{1}{9} = 0 + \frac{4}{9} + \frac{2}{9} = \frac{6}{9} = \frac{2}{3}$$ 4. **Calculate $E[Y]$:** Find marginal $P_Y(y)$ by summing over $X$: - $P_Y(0) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9}$ - $P_Y(1) = \frac{2}{9} + \frac{2}{9} + 0 = \frac{4}{9}$ - $P_Y(2) = \frac{1}{9} + 0 + 0 = \frac{1}{9}$ Then: $$E[Y] = 0 \times \frac{4}{9} + 1 \times \frac{4}{9} + 2 \times \frac{1}{9} = 0 + \frac{4}{9} + \frac{2}{9} = \frac{6}{9} = \frac{2}{3}$$ 5. **Calculate $E[XY]$:** $$E[XY] = \sum_x \sum_y xy P(x,y)$$ Calculate each term: - $(0)(0) \times \frac{1}{9} = 0$ - $(0)(1) \times \frac{2}{9} = 0$ - $(0)(2) \times \frac{1}{9} = 0$ - $(1)(0) \times \frac{2}{9} = 0$ - $(1)(1) \times \frac{2}{9} = \frac{2}{9}$ - $(1)(2) \times 0 = 0$ - $(2)(0) \times \frac{1}{9} = 0$ - $(2)(1) \times 0 = 0$ - $(2)(2) \times 0 = 0$ Sum: $$E[XY] = \frac{2}{9}$$ 6. **Calculate covariance:** $$\mathrm{Cov}(X,Y) = E[XY] - E[X]E[Y] = \frac{2}{9} - \left(\frac{2}{3} \times \frac{2}{3}\right) = \frac{2}{9} - \frac{4}{9} = -\frac{2}{9}$$ 7. **Interpretation:** The covariance is negative, $-\frac{2}{9}$, indicating that when $X$ increases, $Y$ tends to decrease on average, showing a negative linear relationship between $X$ and $Y$ in this distribution.