SOLUTION: A discrete random variable has pdf f(x). (a) If f(x) = k(1/2)^x for x= 1, 2, 3, and zero otherwise, find k. (b) Is a function of the form f(x) = k[(1/2)^x

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Question 1203803: A discrete random variable has pdf f(x).
(a) If f(x) = k(1/2)^x for x= 1, 2, 3, and zero otherwise, find k.
(b) Is a function of the form f(x) = k[(1/2)^x - 1/2] for x = 0, 1, 2 a pdf for any k?
Answer by math_tutor2020(3816) (Show Source):
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Part (a)

f(x) = k(1/2)^x
f(1) = k(1/2)^1
f(1) = k/2

f(x) = k(1/2)^x
f(2) = k(1/2)^2
f(2) = k/4

f(x) = k(1/2)^x
f(3) = k(1/2)^3
f(3) = k/8

For any PDF, the f(x) probability values add to 1.

f(1) + f(2) + f(3) = 1
k/2 + k/4 + k/8 = 1
4k/8 + 2k/8 + k/8 = 1
(4k+2k+k)/8 = 1
7k/8 = 1
7k = 1*8
7k = 8
k = 8/7

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Part (b)

f(x) = k[(1/2)^x - 1/2]
f(0) = k[(1/2)^0 - 1/2]
f(0) = k/2

f(x) = k[(1/2)^x - 1/2]
f(1) = k[(1/2)^1 - 1/2]
f(1) = 0

f(x) = k[(1/2)^x - 1/2]
f(2) = k[(1/2)^2 - 1/2]
f(2) = -k/4

f(0)+f(1)+f(2) = 1
k/2 + 0 + (-k/4) = 1
2k/4 - k/4 = 1
(2k-k)/4 = 1
k/4 = 1
k = 4

f(x) is a PDF if and only if k = 4

SOLUTION: A discrete random variable has pdf f(x). (a) If f(x) = k(1/2)^x for x= 1, 2, 3, and zero otherwise, find k. (b) Is a function of the form f(x) = k[(1/2)^x - 1/2] for x = 0, 1, 2