Problem 3. Probability Density Functions
2 points possible (graded, results hidden)
For t∈R, define the following two functions:
f1(t)=1/√2πexp(−max(1,t^2)^2)
and
f2(t)=1/√2πexp(−min(1,t^2)^2).
In this problem, we explore whether these functions are valid probability density functions.
Determine whether the function f1 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 cf1 is a valid PDF?
Yes, it is a valid PDF.
No, it is not a valid PDF, but there is a constant c making cf1 a valid PDF.
No, it is not a valid PDF, and there is no constant c making cf1 a valid PDF.
None of the above.
unanswered
Determine whether the function f2 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 cf2 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 a valid PDF.
No, it is not a valid PDF, and there is no constant c making cf2 a valid PDF.
None of the above.
unanswered
So, what is the main characteristic of a PDF?
Correct - the area under the curve is 1.
See
https://www.jiskha.com/questions/1798820/For-t-R-define-the-following-two-functions-f1-t-12-exp-max-1-t2-2
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To determine whether a function is a valid Probability Density Function (PDF), we need to check two conditions:
1. The function must be non-negative for all possible values of t.
2. The integral of the function over its entire domain must be equal to 1.
Let's analyze each function separately:
1. For f1(t) = 1/√2π * exp(−max(1,t^2)^2):
- The exponential term is always positive, so we need to determine whether 1/√2π * exp(−max(1,t^2)^2) ≥ 0 for all possible values of t.
When t ≤ -1 or t ≥ 1, max(1,t^2) = 1, and exp(−max(1,t^2)^2) = exp(−1^2) = e^(-1) > 0. So, 1/√2π * exp(−max(1,t^2)^2) ≥ 0 for these values of t.
When -1 < t < 1, max(1,t^2) = t^2, and exp(−max(1,t^2)^2) = exp(−t^4). Since t^4 ≥ 0, exp(−t^4) > 0 for all t in this range.
Therefore, f1(t) is non-negative for all t.
- Next, we need to calculate the integral of f1(t) over its entire domain:
∫[-∞,∞] 1/√2π * exp(−max(1,t^2)^2) dt
The integral of the function cannot be determined analytically, but we can see that it converges because the exponential term decreases rapidly as t increases.
However, since it is not possible to calculate the exact value of the integral, we cannot conclude whether it is equal to 1 or not.
Thus, we cannot determine if f1(t) is a valid PDF.
2. For f2(t) = 1/√2π * exp(−min(1,t^2)^2):
- Similar to f1(t), the exponential term is always positive. We need to determine whether 1/√2π * exp(−min(1,t^2)^2) ≥ 0 for all possible values of t.
When t ≤ -1 or t ≥ 1, min(1,t^2) = t^2, and exp(−min(1,t^2)^2) = exp(−t^4) > 0.
When -1 < t < 1, min(1,t^2) = 1, and exp(−min(1,t^2)^2) = exp(−1^2) = e^(-1) > 0.
Therefore, f2(t) is non-negative for all t.
- We again need to calculate the integral of f2(t) over its entire domain:
∫[-∞,∞] 1/√2π * exp(−min(1,t^2)^2) dt
Similar to f1(t), it is not possible to calculate the exact value of this integral analytically. However, we can observe that the exponential term decreases rapidly as t increases.
Again, we cannot determine if f2(t) is a valid PDF based on this information.
Therefore, the correct answers to the questions are:
- For f1(t): No, it is not a valid PDF, but there is no constant c making cf1 a valid PDF.
- For f2(t): No, it is not a valid PDF, but there is no constant c making cf2 a valid PDF.
To determine if a function is a valid probability density function (PDF), we need to check two conditions:
1. The function must be non-negative: For any value of t, the function f(t) must be greater than or equal to 0.
2. The integral of the function over the entire range must be equal to 1: ∫f(t) dt = 1.
Let's analyze each function separately:
1. Function f1(t) = 1/√2πexp(−max(1,t^2)^2):
To check the first condition, we need to ensure that f1(t) is non-negative for all real numbers t. The exponential term exp(−max(1,t^2)^2) is always positive, so f1(t) will be non-negative as long as 1/√2π is positive. Since 1/√2π is a positive constant, f1(t) is non-negative.
Next, we need to check if the integral of f1(t) over the entire real line is equal to 1:
∫f1(t) dt = ∫(1/√2πexp(−max(1,t^2)^2)) dt
To evaluate this integral, we need to split it into two parts:
∫(1/√2πexp(−max(1,t^2)^2)) dt = ∫(1/√2πexp(−t^4)) dt for t ≤ -1
+ ∫(1/√2πexp(−1)) dt for -1 < t < 1
+ ∫(1/√2πexp(−t^4)) dt for t ≥ 1
The integral of the first and third parts may not converge since the exponential term exp(−t^4) decays very slowly for large positive or negative t values. Therefore, the integral of f1(t) cannot be equal to 1.
Hence, f1(t) is not a valid PDF.
However, we can multiply f1(t) by a positive constant c to make the integral equal to 1:
∫(cf1(t)) dt = c∫(1/√2πexp(−max(1,t^2)^2)) dt
So, there exists a constant c > 0 such that cf1(t) is a valid PDF.
Therefore, the correct answer for the first part is:
No, it is not a valid PDF, but there is a constant c making cf1 a valid PDF.
2. Function f2(t) = 1/√2πexp(−min(1,t^2)^2):
To check the first condition, we need to ensure that f2(t) is non-negative for all real numbers t. Similar to f1(t), the exponential term exp(−min(1,t^2)^2) is always positive. Hence, f2(t) will be non-negative as long as 1/√2π is positive. Since 1/√2π is a positive constant, f2(t) is non-negative.
Next, we need to check if the integral of f2(t) over the entire real line is equal to 1:
∫f2(t) dt = ∫(1/√2πexp(−min(1,t^2)^2)) dt
To evaluate this integral, we need to split it into two parts:
∫(1/√2πexp(−min(1,t^2)^2)) dt = ∫(1/√2πexp(−1)) dt for t ≤ -1
+ ∫(1/√2πexp(−t^4)) dt for -1 < t < 1
+ ∫(1/√2πexp(−t^4)) dt for t ≥ 1
Unlike f1(t), the integrals in f2(t) are well-behaved. The integral of the first and third parts will converge since the exponential term exp(−t^4) decays rapidly for large positive or negative t values. The second part is a constant.
Therefore, the integral of f2(t) will be equal to 1.
Hence, f2(t) is a valid PDF.
The correct answer for the second part is:
Yes, it is a valid PDF.
Therefore, the overall answer is:
No, it is not a valid PDF, but there is a constant c making cf1 a valid PDF.
Yes, it is a valid PDF.