SOLUTION: A discrete random variable X is such that P(X = n) = 2^n−1 / 3^n , n = 1, 2, . . . , n, . . . . Show that EX = 3. Thank You

Question 1177293: A discrete random variable X is such that
P(X = n) = 2^n−1 / 3^n , n = 1, 2, . . . , n, . . . .
Show that EX = 3.
Thank You
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**1. Define the Expected Value**
The expected value of a discrete random variable X is defined as:
* E(X) = Σ [x * P(X = x)]
In our case, X takes values n, and P(X = n) = 2^(n-1) / 3^n. So:
* E(X) = Σ [n * (2^(n-1) / 3^n)] for n = 1, 2, 3, ...
**2. Simplify the Expression**
* E(X) = Σ [n * (2^(n-1) / 3^n)]
* E(X) = Σ [n * (1/2) * (2^n / 3^n)]
* E(X) = (1/2) * Σ [n * (2/3)^n]
**3. Recognize the Series**
We have the series:
* Σ [n * (2/3)^n]
Let's use the formula for the sum of the series Σ nx^(n-1) = 1/(1-x)^2
We can rewrite the sum as:
* Σ [n * (2/3)^n] = (2/3) * Σ [n * (2/3)^(n-1)]
Now, let x = 2/3. Then:
* Σ [n * (2/3)^(n-1)] = 1 / (1 - 2/3)^2
* Σ [n * (2/3)^(n-1)] = 1 / (1/3)^2
* Σ [n * (2/3)^(n-1)] = 1 / (1/9)
* Σ [n * (2/3)^(n-1)] = 9
Therefore:
* Σ [n * (2/3)^n] = (2/3) * 9 = 6
**4. Calculate E(X)**
* E(X) = (1/2) * Σ [n * (2/3)^n]
* E(X) = (1/2) * 6
* E(X) = 3
**Therefore, E(X) = 3.**

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