For t element of R, define the following two functions:

Question

f_1 (t)=1/sqrt(2*pi)*e^(-max(1,t^2)/2) and f_2 (t)=1/sqrt(2*pi)*e^(-min(1,t^2)/2)

In this problem, we explore whether these functions are valid probability density functions.

1. Determine whether the function f_1 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 c*f_1 is a valid PDF?
a. Yes, it a valid PDF
b. No, its not a valid PDF, but there is a constant c making c*f_1 a valid PDF
c. No, it is not a valid PDF, and there is no constant c making c*f_1 a valid PDF.
d. None of the above.

2. Determine whether the function f_2 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 c*f_2 is a valid PDF?
a. Yes, it a valid PDF
b. No, its not a valid PDF, but there is a constant c making c*f_2 a valid PDF
c. No, it is not a valid PDF, and there is no constant c making c*f_2 a valid PDF.
d. None of the above.

Answer 1

To determine whether a function is a valid probability density function (PDF), it needs to satisfy two conditions:

1. The function must be non-negative for all values of t.
2. The integral of the function over its entire support (in this case, the real line) must equal 1.

Let's analyze each function separately:

1. f_1(t) = 1/sqrt(2*pi)*e^(-max(1,t^2)/2)

First, note that e^(-max(1,t^2)/2) is always positive for any value of t, so we only need to focus on the multiplying factor 1/sqrt(2*pi). This factor is a constant and does not change with t. Since it is positive, it does not affect the positivity of the function.

Now, let's look at the integral of f_1(t):

∫(from -∞ to ∞) (1/sqrt(2*pi)*e^(-max(1,t^2)/2)) dt

Note that e^(-max(1,t^2)/2) decays rapidly as t increases, and approaches 0 as t approaches ±∞. Therefore, the integral of f_1(t) over the entire real line will be finite and positive, but it will not equal 1. Hence, f_1(t) is not a valid PDF.

To determine if there exists a constant c > 0 such that c*f_1(t) is a valid PDF, we need to find a value of c that makes the integral of c*f_1(t) equal to 1. However, no matter what value of c we choose, the integral will still not equal 1 because the shape of the function cannot be changed by scaling it with a constant. Therefore, the answer to question 1 is:

c. No, it is not a valid PDF, and there is no constant c making c*f_1 a valid PDF.

2. f_2(t) = 1/sqrt(2*pi)*e^(-min(1,t^2)/2)

Similarly to f_1(t), e^(-min(1,t^2)/2) is always positive for any value of t. The multiplying factor 1/sqrt(2*pi) is also always positive and does not change with t. Hence, the positivity condition is satisfied.

Now, let's consider the integral of f_2(t):

∫(from -∞ to ∞) (1/sqrt(2*pi)*e^(-min(1,t^2)/2)) dt

The function e^(-min(1,t^2)/2) also decays rapidly as t increases, and approaches 0 as t approaches ±∞. Therefore, the integral of f_2(t) over the entire real line will be finite and positive, but it will not equal 1. Hence, f_2(t) is not a valid PDF.

Similarly to the previous question, no constant c > 0 can be chosen to make the integral of c*f_2(t) equal to 1, as the shape of the function cannot be changed by scaling it with a constant. Therefore, the answer to question 2 is:

c. No, it is not a valid PDF, and there is no constant c making c*f_2 a valid PDF.

So, the overall answer is:

c. No, it is not a valid PDF, and there is no constant c making c*f_1 a valid PDF. No, it is not a valid PDF, and there is no constant c making c*f_2 a valid PDF.