Questions
Multiple choice
Question at position 2 Which of the following are valid probability functions? Select all that apply. Note that, in this question, f(x)f(x) indicates a continuous probability distribution function, while P(X=x)P(X=x) indicates a discrete probability distribution function.f(x)=x,x∈[−1,1]f(x) = x, x \in [-1,1]P(X=x)=x15,x∈{0,1,2,3,4,5}P(X=x) = \frac{x}{15}, x \in \{0,1,2,3,4,5\}P(X=x)=14,x∈{0,1,2,3,4}P(X=x) = \frac{1}{4}, x \in \{0,1,2,3,4 \}P(X=x)=x15,x∈{1,2,3,4,5}P(X=x) = \frac{x}{15}, x \in \{1,2,3,4,5\}f(x)=1,x∈[0,2]f(x) = 1, x \in [0,2]f(x)=1−x2,x∈[0,2]f(x) = 1 - \frac{x}{2}, x \in [0,2]P(X=x)=12,x∈{0,1}P(X=x) = \frac{1}{2}, x \in \{0,1\}
Options
A.f
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B.P
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C.P
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G.P
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View Explanation
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Step-by-Step Analysis
Let's take each candidate in turn and check whether it satisfies the properties of a probability function.
Option 1: f(x) = x for x in [-1, 1]. A continuous probability density must be nonnegative everywhere on its support and integrate to 1. Here f(x) is negative for x < 0, since x < 0 in part of the interval. That immediately disqualifies it as a valid density. Moreover, even if we restrict to nonnegative parts, the integral over [-1,1] would......Login to view full explanationLog in for full answers
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