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.
To determine if the function f₁ is a valid PDF, we need to check if its integral over the entire real line is equal to 1.
∫f₁(t) dt = ∫(1/√(2π)) * exp((-max(1,t²))/2) dt.
Since the function is defined differently for negative and positive values of t, we can split the integral into two parts:
∫f₁(t) dt = ∫(1/√(2π)) * exp((-max(1,t²))/2) dt from -∞ to -1 + ∫(1/√(2π)) * exp((-max(1,t²))/2) dt from 1 to ∞.
For the first integral, let t' = -t. Then the integral becomes:
∫(1/√(2π)) * exp((-max(1,t²))/2) dt' from -∞ to -1.
Since exp((-max(1,t²))/2) is always positive, we can remove the max function:
∫(1/√(2π)) * exp(-t'²/2) dt' from -∞ to -1.
This is the integral of a standard normal distribution from -∞ to -1, which is equal to the probability that a standard normal random variable is less than -1, P(Z < -1). This probability is positive but less than 1.
For the second integral, we can remove the max function as well:
∫(1/√(2π)) * exp(-t²/2) dt from 1 to ∞.
This is the integral of a standard normal distribution from 1 to ∞, which is equal to the probability that a standard normal random variable is greater than 1, P(Z > 1). This probability is positive but less than 1.
Since both of these integrals are less than 1, the total integral over the entire real line is less than 2. Therefore, f₁ is not a valid PDF because its integral is not equal to 1.
To determine if there is a constant c>0 such that cf₁ is a valid PDF, we need to find the value of c that makes the integral of cf₁ equal to 1.
Let's consider the integral of cf₁:
∫cf₁(t) dt = ∫c(1/√(2π)) * exp((-max(1,t²))/2) dt.
We can split this integral into two parts as well:
∫cf₁(t) dt = ∫c(1/√(2π)) * exp((-max(1,t²))/2) dt from -∞ to -1 + ∫c(1/√(2π)) * exp((-max(1,t²))/2) dt from 1 to ∞.
The integrals in both parts are equal to c*P(Z < -1) and c*P(Z > 1) respectively, where Z is a standard normal random variable. Since both of these probabilities are positive but less than 1, there is no value of c>0 that can make the integral equal to 1.
Therefore, f₁ is not a valid PDF, and there is no constant c>0 that can make cf₁ a valid PDF.
The correct answer is: No, it is not a valid PDF, and there is no constant c making cf₁ a valid PDF.
To determine whether the function f_1 is a valid probability density function (PDF) for a continuous random variable that can take any value on the real line, we need to ensure that the following conditions are satisfied:
1. The function f_1(t) is non-negative for all t in ℝ.
2. The integral of f_1(t) over the entire real line (∫_{-∞}^{∞} f_1(t) dt) is equal to 1.
First, let's check if f_1(t) is non-negative for all t in ℝ:
f_1(t) = (1/√(2π)) * exp(-(max(1, t^2))/2)
Since both √(2π) and exp(-(max(1, t^2))/2) are always non-negative, the overall function f_1(t) also remains non-negative.
Now, let's calculate the integral of f_1(t) over the entire real line:
∫_{-∞}^{∞} f_1(t) dt = ∫_{-∞}^{∞} (1/√(2π)) * exp(-(max(1, t^2))/2) dt
To evaluate this integral, we need to consider the cases where t^2 is either less than or equal to 1, or greater than 1.
Case 1: If t^2 ≤ 1, then max(1, t^2) = 1, and the integral becomes:
∫_{-∞}^{∞} (1/√(2π)) * exp(-1/2) dt
= (1/√(2π)) * exp(-1/2) * ∫_{-∞}^{∞} dt
= (1/√(2π)) * exp(-1/2) * ∞
= ∞
Case 2: If t^2 > 1, then max(1, t^2) = t^2, and the integral becomes:
∫_{-∞}^{∞} (1/√(2π)) * exp(-t^2/2) dt
This integral equals √(2π), which can be shown using standard methods of integration. Therefore, we have:
∫_{-∞}^{∞} f_1(t) dt = √(2π)
Since the integral does not converge to 1, f_1(t) is not a valid PDF for a continuous random variable that can take any value on the real line.
However, there is a number c > 0 such that c * f_1 is a valid PDF. To find this constant, we need to ensure that the integral of c * f_1(t) over the entire real line is equal to 1.
Let's rewrite c * f_1(t) as follows:
c * f_1(t) = c * (1/√(2π)) * exp(-(max(1, t^2))/2)
Now, we need to determine the value of c that makes the integral of c * f_1(t) over the entire real line equal to 1.
Let's consider the two cases again:
Case 1: If t^2 ≤ 1, then max(1, t^2) = 1, and the integral becomes:
∫_{-∞}^{∞} c * (1/√(2π)) * exp(-1/2) dt
= c * (1/√(2π)) * exp(-1/2) * ∫_{-∞}^{∞} dt
= c * (1/√(2π)) * exp(-1/2) * ∞
= ∞
Case 2: If t^2 > 1, then max(1, t^2) = t^2, and the integral becomes:
∫_{-∞}^{∞} c * (1/√(2π)) * exp(-t^2/2) dt
This integral equals c * √(2π), which can be shown using standard methods of integration. Therefore, we have:
c * √(2π) = 1
To make the integral equal to 1, we need to set c = 1/√(2π).
So, we have found a constant c = 1/√(2π) such that c * f_1 is a valid PDF.
Therefore, the correct answer is:
No, it is not a valid PDF, but there is a constant c > 0 making cf_1 a valid PDF.