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.
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.
To determine whether a function is a valid probability density function (PDF), we need to check if it satisfies two conditions:
1. The function must always be non-negative.
2. The area under the function's curve (i.e., the integral) must equal 1.
Let's analyze each function separately:
For f_1(t) = (1/√(2π)) * exp(-max(1, t^2)/2):
1. The exponential term exp(-max(1, t^2)/2) is always non-negative.
2. However, the function f_1(t) is not always non-negative because of the max(1, t^2) term. When t is less than -1 or greater than 1, the function becomes negative, which violates the first condition.
Therefore, the function f_1(t) is not a valid PDF in its current form.
Now, let's analyze f_2(t) = (1/√(2π)) * exp(-min(1, t^2)/2):
1. The exponential term exp(-min(1, t^2)/2) is always non-negative.
2. The function f_2(t) is also always non-negative since the min(1, t^2) term does not affect its sign.
To confirm whether f_2(t) is a valid PDF, we need to check if its integral over the entire real line equals 1. However, integrating f_2(t) over the real line is not possible due to its non-elementary form. Thus, we cannot determine if f_2(t) is a valid PDF directly through integration.
However, we can make use of the fact that the exponential term decreases rapidly as the argument increases. By observing that t^2 is always positive or zero, it follows that min(1, t^2) is always less than or equal to 1. Therefore, f_2(t) will be largest when t^2 = 0, which occurs when t = 0.
Hence, by considering the limiting behavior of f_2(t) as t approaches infinity or negative infinity, we can conclude that the integral of f_2(t) will converge and be equal to some positive number.
Thus, we can conclude that f_2(t) is a valid PDF.