Discrete random variables and PMF

[Red88]

Homework Statement

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.

Homework 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)

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

 

[mighty2000]

Dear Red88,

Thank you for posting your question on the forum. I am a scientist and I would be happy to help you with your homework problem.

First, let's address the question of what values of "a" are allowed in this PMF. As you correctly stated, each p(xi) must be greater than or equal to zero and the sum of all p(xi) must equal 1. This means that "a" must satisfy the following conditions:

1-a >= 0
1-2a >= 0
0.2 >= 0
a >= 0
0.5a >= 0
1-a + 1-2a + 0.2 + a + 0.5a = 1

Solving these inequalities and equation, we get:

0 <= a <= 0.5

Therefore, the allowed values of "a" in this PMF are between 0 and 0.5, inclusive.

Now, let's calculate the expected value and standard deviation of the variable X. You are on the right track with your calculations, but there is a small mistake in the calculation of the variance. The correct formula for variance is:

V(x) = E(x^2) - [E(x)]^2

Using this formula, we get:

E(x) = (1)(1-a) + (2)(1-2a) + (3)(0.2) + (4)(a) + (5)(0.5a) = 4.8 - 3.5a
E(x^2) = (1)^2(1-a) + (2)^2(1-2a) + (3)^2(0.2) + (4)^2(a) + (5)^2(0.5a) = 22.4 - 11a
V(x) = 22.4 - 11a - (4.8 - 3.5a)^2 = 22.4 - 11a - (23.04 - 26.4a + 12.25a^2) = -0.64 + 15.9a - 12.25a^2

Now, we can use this expression for variance to calculate the standard deviation:

SD = √V(x) = √(-0.64 + 15.9a - 12.25a