For t element of R, define the following two functions:
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.
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.
To determine whether a function is a valid probability density function (PDF), we need to check if the function satisfies two conditions:
1. The function is non-negative for all values of the random variable.
2. The area under the curve of the function over the entire range of possible values is equal to 1.
Let's analyze each function separately:
1. For the function f_1(t) = 1/sqrt(2*pi) * e^(-max(1, t^2)/2):
a. Is the function non-negative for all t?
Yes, because 1/sqrt(2*pi) is a positive constant, and e^(-max(1, t^2)/2) is always positive or zero.
b. Does the function integrate to 1 over the entire range of possible values of t?
No, because the function decays and approaches 0 as t approaches infinity, but it does not approach 0 fast enough to integrate to 1 over the entire real line.
Therefore, f_1(t) is not a valid PDF.
However, we can consider the constant c = 1/2. If we multiply f_1(t) by c, we get c*f_1(t) = (1/2)*(1/sqrt(2*pi)) * e^(-max(1, t^2)/2), which is also a non-negative function.
c. Is c*f_1(t) a valid PDF?
No, because even though it is non-negative, the integral of c*f_1(t) over the entire real line is not equal to 1.
So, the answer is (c) No, it is not a valid PDF, and there is no constant c making c*f_1 a valid PDF.
2. For the function f_2(t) = 1/sqrt(2*pi) * e^(-min(1, t^2)/2):
a. Is the function non-negative for all t?
Yes, because 1/sqrt(2*pi) is a positive constant, and e^(-min(1, t^2)/2) is always positive or zero.
b. Does the function integrate to 1 over the entire range of possible values of t?
No, because the function decays and approaches 0 as t approaches minus infinity, but it does not approach 0 fast enough to integrate to 1 over the entire real line.
Therefore, f_2(t) is not a valid PDF.
Again, let's try to find a constant c that makes c*f_2(t) a valid PDF. In this case, we can choose any positive constant c, because multiplying by a positive constant does not change the non-negativity property.
c. Is c*f_2(t) a valid PDF for any positive constant c?
No, because even though it is non-negative, the integral of c*f_2(t) over the entire real line is not equal to 1.
So, the answer is (c) No, it is not a valid PDF, and there is no constant c making c*f_2 a valid PDF.
In conclusion:
1. The function f_1(t) is not a valid PDF, and there is no constant c making c*f_1 a valid PDF. Therefore, the answer is (c).
2. The function f_2(t) is not a valid PDF, and there is no constant c making c*f_2 a valid PDF. Therefore, the answer is (c).