Question 1202002: Karlson sends a message to Mažylis consisting of two bits in length. When he is in a good mood (with probability q), he sends '11', otherwise he sends '01'. Each bit is transmitted over the communication channel with a probability of distortion p. Let X be the number of ones in the message sent by Karlson and Y be the number of ones in the message received by Mažylis. Find the covariance between X and Y.
(p = 0.26, q = 0.58)
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **1. Determine the Probability Mass Function (PMF) of X**
* **Good Mood (Probability q = 0.58):** Sends "11"
* P(X=2) = 0.58
* **Bad Mood (Probability 1-q = 0.42):** Sends "01"
* P(X=1) = 0.42
**2. Determine the Possible Received Messages (Y) and their Probabilities**
* **"11" Sent:**
* Received "11": Probability = 0.58 * (1-p) * (1-p) = 0.58 * 0.74 * 0.74 = 0.3175
* Received "10": Probability = 0.58 * (1-p) * p = 0.58 * 0.74 * 0.26 = 0.1129
* Received "01": Probability = 0.58 * p * (1-p) = 0.58 * 0.26 * 0.74 = 0.1129
* Received "00": Probability = 0.58 * p * p = 0.58 * 0.26 * 0.26 = 0.0392
* **"01" Sent:**
* Received "11": Probability = 0.42 * p * (1-p) = 0.42 * 0.26 * 0.74 = 0.0808
* Received "10": Probability = 0.42 * p * p = 0.42 * 0.26 * 0.26 = 0.0283
* Received "01": Probability = 0.42 * (1-p) * p = 0.42 * 0.74 * 0.26 = 0.0808
* Received "00": Probability = 0.42 * (1-p) * (1-p) = 0.42 * 0.74 * 0.74 = 0.2289
**3. Calculate E[X], E[Y], and E[XY]**
* **E[X] (Expected value of X):**
* E[X] = (2 * 0.58) + (1 * 0.42) = 1.58
* **E[Y]:**
* E[Y] = (2 * 0.3175) + (1 * 0.1129 + 0.1129 + 0.0808 + 0.0808) + (0 * 0.0392 + 0.0283 + 0.2289)
* E[Y] = 0.635 + 0.3874 + 0.2572 = 1.2796
* **E[XY]:**
* E[XY] = (2 * 2 * 0.3175) + (2 * 1 * 0.1129) + (1 * 2 * 0.0808) + (1 * 1 * 0.0808) + (0 * 2 * 0.0392) + (0 * 1 * 0.0283) + (0 * 1 * 0.2289)
* E[XY] = 1.27 + 0.2258 + 0.1616 + 0.0808
* E[XY] = 1.7382
**4. Calculate Covariance**
* **Cov(X, Y) = E[XY] - E[X] * E[Y]**
* Cov(X, Y) = 1.7382 - (1.58 * 1.2796)
* Cov(X, Y) = 1.7382 - 2.0215
* Cov(X, Y) = -0.2833
**Therefore, the covariance between X (number of ones sent) and Y (number of ones received) is -0.2833.**
This negative covariance indicates that there is a tendency for the number of ones in the sent message to be inversely related to the number of ones in the received message.
|
|
|
