For t∈R, define the following two functions:
f1(t)=(1/√2π)*exp(−max(1,t^2)/2)
and
f2(t)=(1√2π)*exp(−min(1,t^2)/2).
In this problem, we explore whether these functions are valid probability density functions.
1. 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?
Choose the correct answer:
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.
2. 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?
Choose the correct answer:
Yes, it is a valid PDF.
No, it is not a valid PDF, but there is a constant c making cf2 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.
1. No, it is not a valid PDF, but there is a constant c making cf1 a valid PDF.
To determine whether f1 is a valid PDF, we need to check if it satisfies the properties of a PDF:
1. The function must be non-negative for all values of t. The exponential function is always positive, so exp(-max(1,t^2)/2) is positive for all values of t. Multiplying by (1/√(2π)) doesn't affect the positivity, so f1(t) is non-negative.
2. The integral of the function over the entire real line must be equal to 1. However, the function f1(t) does not integrate to 1 over the entire real line because max(1, t^2) becomes 1 for t > 1. Therefore, f1 is not a valid PDF.
However, we can multiply f1 by a constant c to make it a valid PDF. We need to find the value of c that makes c*f1 integrate to 1. Since f1 does integrate to a constant value over t in (-∞, 1], we can set c to be the reciprocal of that constant value. Therefore, there exists a constant c > 0 such that cf1 is a valid PDF.
2. No, it is not a valid PDF, but there is a constant c making cf2 a valid PDF.
To determine whether f2 is a valid PDF, we need to check if it satisfies the properties of a PDF:
1. The function must be non-negative for all values of t. Similar to f1, exp(-min(1,t^2)/2) is positive for all values of t, so f2(t) is non-negative.
2. The integral of the function over the entire real line must be equal to 1. However, the function f2(t) does not integrate to 1 over the entire real line because min(1, t^2) becomes 1 for t < -1 or t > 1. Therefore, f2 is not a valid PDF.
Similarly to f1, we can multiply f2 by a constant c to make it a valid PDF. We need to find the value of c that makes c*f2 integrate to 1. Since f2 does integrate to a constant value over t in [-1, 1], we can set c to be the reciprocal of that constant value. Therefore, there exists a constant c > 0 such that cf2 is a valid PDF.
To determine whether the functions f1(t) and f2(t) are valid probability density functions (PDFs), we need to check two conditions:
1. The function must be non-negative for all values of t.
2. The integral of the function over its entire domain must be equal to 1.
Let's check these conditions for f1(t):
1. f1(t) = (1/√2π) * exp(-max(1,t^2)/2)
Since the exponential function is always positive, f1(t) will always be non-negative for all values of t.
2. To check the integral, we need to calculate ∫f1(t) dt over its entire domain. Since the function is defined for all real numbers, the domain is (-∞, ∞).
∫f1(t) dt = ∫(1/√2π) * exp(-max(1,t^2)/2) dt
This integral is complicated to solve analytically, but we can conclude that if we integrate over the entire real line, the result will be a positive finite number. However, we cannot determine if it equals 1 without further computation. Therefore, we cannot definitively determine if f1(t) is a valid PDF.
For f2(t):
1. f2(t) = (1/√2π) * exp(-min(1,t^2)/2)
Similar to f1(t), f2(t) will always be non-negative for all values of t.
2. Let's calculate the integral of f2(t):
∫f2(t) dt = ∫(1/√2π) * exp(-min(1,t^2)/2) dt
Again, this integral is complicated to solve analytically, but we can conclude that if we integrate over the entire real line, the result will be a positive finite number. However, we cannot determine if it equals 1 without further computation. Therefore, we cannot definitively determine if f2(t) is a valid PDF either.
In conclusion:
1. The answer for f1(t) is: No, it is not a valid PDF, and there is no constant c making cf1 a valid PDF.
2. The answer for f2(t) is: No, it is not a valid PDF, and there is no constant c making cf2 a valid PDF.