Discrete random variables and PMF

  • Thread starter Red88
  • Start date
1. The problem statement, all variables and given/known data

A discrete random variable X has the following PMF

x | 1 | 2 | 3 | 4 | 5 |
p(x)|1-a|1-2a|0.2| a | 0.5a|

What are the values of "a" that are allowed in this PMF?
For the allowed values, compute the expected value and the standard deviation of the variable X.


2. Relevant equations

Each p(xi) >= 0 and the sum of all p(xi) from i = 1 to i = 5 is equal to 1
E(x) = x1(p(x1)) + x2(p(x2)) + x3(p(x3)) + x4(p(x4)) + x5(p(x5))
E(x^2) = x1^2(p(x1)) + x2^2(p(x2)) + x3^2(p(x3)) + x4^2(p(x4)) + x5^2(p(x5))
V(x) = E(x^2) - [E(x)]^2
SD (standard deviation) = sq. root of V(X)


3. The attempt at a solution

(1-a) + (1-2a) + 0.2 + a + 0.5a = 1
2.2 - 1.5a = 1 => a = 0.8

E(x) = (1)(1-(.8)) + (2)(1-2(.8)) + (3)(0.2) + (4)(0.8) + (5)(.5 * .8) = 4.8
E(x^2) = (1)^2(1-(.8)) + (2)^2(1-2(.8)) + (3)^2(0.2) + (4)^2(0.8) + (5)^2(.5 * .8) = 22.4
V(x) = 22.4 - [4.8]^2 => V(x) = -0.64, which is not possible since variance is nonnegative or else we would need to take the sq. root of a negative number to get SD, which can only be a real, positive value......

Any help would be greatly appreciated since I need to turn in this HW by Monday!

-Red88
 

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