**Step 1:** Start with the definition of mutual information using the entropies: \[ I(x; Y) = H(x) - H(x|Y) \] Recall from the definitions: \[ H(x) = -\sum_x P(x) \log_2 P(x) \] \[ H(x|Y) = -\sum_{x, y} P(x, y) \log_2 \left(\frac{P(x, y)}{P(y)}\right) \] **Explanation:** We are expressing mutual information in terms of entropy \( H(x) \) and conditional entropy \( H(x|Y) \), as suggested in the hint. --- **Step 2:** If \( x \) and \( Y \) are statistically independent, then \( P(x, y) = P(x)P(y) \). Plug this into the formula for \( H(x|Y) \): \[ H(x|Y) = -\sum_{x, y} P(x, y) \log_2 \left(\frac{P(x, y)}{P(y)}\right) = -\sum_{x, y} P(x)P(y) \log_2 \left(\frac{P(x)P(y)}{P(y)}\right) = -\sum_{x, y} P(x)P(y) \log_2 P(x) \] Since \( \sum_y P(y) = 1 \): \[ H(x|Y) = -\sum_x P(x) \log_2 P(x) \sum_y P(y) = -\sum_x P(x) \log_2 P(x) = H(x) \] Therefore, \[ I(x; Y) = H(x) - H(x|Y) = H(x) - H(x) = 0 \] **Explanation:** When \( x \) and \( Y \) are independent, knowing \( Y \) gives no information about \( x \), so the conditional entropy \( H(x|Y) \) equals the entropy \( H(x) \), proving mutual information is zero. --- **Final Answer:** \[ \boxed{I(x; Y) = 0 \text{ if } x \text{ and } Y \text{ are statistically independent.} } \]