The function f_1 is not a valid PDF for a continuous random variable that can take any value on the real line. This is because the numerator, e^(-max(1,t^2)/2), can take values greater than 1 for certain values of t, which violates the condition that a PDF must have a maximum value of 1. Therefore, there is no constant c>0 that can make cf_1 a valid PDF.
The function f_2 is a valid PDF for a continuous random variable that can take any value on the real line. This is because the numerator, e^(-min(1,t^2)/2), is always less than or equal to 1 for all values of t. Therefore, there is no need to multiply f_2 by a constant c to make it a valid PDF. Thus, the answer is: Yes, f_2 is a valid PDF.
For t\in \mathbb {R}, define the following two functions:
f_1(t)=\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{\max (1,t^2)}{2}\right)
and
f_2(t)=\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{\min (1,t^2)}{2}\right).
In this problem, we explore whether these functions are valid probability density functions.
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?
Yes, it is a valid PDF.
No, it is not a valid PDF, but there is a constant c making cf_1 a valid PDF.
No, it is not a valid PDF, and there is no constant c making cf_1 a valid PDF.
None of the above.
submitted
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?
Yes, it is a valid PDF.
No, it is not a valid PDF, but there is a constant c making cf_2 a valid PDF.
No, it is not a valid PDF, and there is no constant c making cf_2 a valid PDF.
None of the above.
1 answer