Question 1198348: A random variable X has support containing only two numbers. Its
expected value is EX = 5. Its variance is Var X = 3. Give an example of the
pmf of such a random variable
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **1. Define the Random Variable**
* Let X be the random variable representing the possible values of the random variable.
* Since X has support containing only two numbers, let's assume those numbers are 'a' and 'b'.
**2. Define the Probability Mass Function (PMF)**
* Let P(X = a) = p
* Let P(X = b) = 1 - p
**3. Set up Equations Based on Given Information**
* **Expected Value (E[X]):**
* E[X] = a * P(X = a) + b * P(X = b)
* 5 = a * p + b * (1 - p)
* **Variance (Var(X)):**
* Var(X) = E[X²] - (E[X])²
* 3 = E[X²] - 5²
* E[X²] = 28
* E[X²] = a² * P(X = a) + b² * P(X = b)
* 28 = a² * p + b² * (1 - p)
**4. Solve the System of Equations**
* We have two equations and two unknowns (a, b, and p).
* Solve these equations simultaneously to find the values of 'a', 'b', and 'p'.
**Example Solution**
* **Let's assume:**
* a = 2
* b = 8
* **Solve for p using E[X] = 5:**
* 5 = 2 * p + 8 * (1 - p)
* 5 = 2p + 8 - 8p
* 6p = 3
* p = 1/2
* **Verify Var(X) = 3:**
* E[X²] = 2² * (1/2) + 8² * (1/2) = 34
* Var(X) = E[X²] - (E[X])² = 34 - 5² = 3
**Therefore, one possible PMF for X is:**
* P(X = 2) = 1/2
* P(X = 8) = 1/2
**Note:** This is just one possible solution. There might be other combinations of 'a', 'b', and 'p' that also satisfy the given conditions for expected value and variance.
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