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.
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.
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