For t∈R, define the following two functions:

Question

f1(t)=(1/√2π)*exp(−max(1,t^2)/2)
 
and

f2(t)=(1√2π)*exp(−min(1,t^2)/2).
 
In this problem, we explore whether these functions are valid probability density functions.

1. Determine whether the function f1 is a valid PDF for a continuous random variable that can take any value on the real line. If not, is there a number c>0, such that cf1 is a valid PDF?
Choose the correct answer: 
Yes, it is a valid PDF.
No, it is not a valid PDF, but there is a constant  c  making  cf1  a valid PDF.
No, it is not a valid PDF, and there is no constant  c  making  cf1  a valid PDF.
None of the above.

2. Determine whether the function f2 is a valid PDF for a continuous random variable that can take any value on the real line. If not, is there a number c>0, such that cf2 is a valid PDF? 
Choose the correct answer:
Yes, it is a valid PDF.
No, it is not a valid PDF, but there is a constant  c  making  cf2  a valid PDF.
No, it is not a valid PDF, and there is no constant  c  making  cf2  a valid PDF.
None of the above.

GPT 3.5 · Answer

1. No, it is not a valid PDF, but there is a constant c making cf1 a valid PDF.

To determine whether f1 is a valid PDF, we need to check if it satisfies the properties of a PDF:
1. The function must be non-negative for all values of t. The exponential function is always positive, so exp(-max(1,t^2)/2) is positive for all values of t. Multiplying by (1/√(2π)) doesn't affect the positivity, so f1(t) is non-negative.
2. The integral of the function over the entire real line must be equal to 1. However, the function f1(t) does not integrate to 1 over the entire real line because max(1, t^2) becomes 1 for t > 1. Therefore, f1 is not a valid PDF.

However, we can multiply f1 by a constant c to make it a valid PDF. We need to find the value of c that makes c*f1 integrate to 1. Since f1 does integrate to a constant value over t in (-∞, 1], we can set c to be the reciprocal of that constant value. Therefore, there exists a constant c > 0 such that cf1 is a valid PDF.

2. No, it is not a valid PDF, but there is a constant c making cf2 a valid PDF.

To determine whether f2 is a valid PDF, we need to check if it satisfies the properties of a PDF:
1. The function must be non-negative for all values of t. Similar to f1, exp(-min(1,t^2)/2) is positive for all values of t, so f2(t) is non-negative.
2. The integral of the function over the entire real line must be equal to 1. However, the function f2(t) does not integrate to 1 over the entire real line because min(1, t^2) becomes 1 for t < -1 or t > 1. Therefore, f2 is not a valid PDF.

Similarly to f1, we can multiply f2 by a constant c to make it a valid PDF. We need to find the value of c that makes c*f2 integrate to 1. Since f2 does integrate to a constant value over t in [-1, 1], we can set c to be the reciprocal of that constant value. Therefore, there exists a constant c > 0 such that cf2 is a valid PDF.