For t∈R, define the following two functions:
f1(t)=12π−−√exp(−max(1,t2)2)
and
f2(t)=12π−−√exp(−min(1,t2)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.
clearly f2 is not a valid PDF, since for |t|>1, f1(t) = k/√e so the area under the curve outside [-1,1] is unbounded.
similarly, f1 is not a PDF, but since it has a limit, a suitable c can be found to make the ∫[-∞,∞] f2(t) dt = 1
a) No, it is not a valid PDF, but there is a constant c making cf1 a valid PDF.
b) No, it is not a valid PDF, and there is no constant c making cf2 a valid PDF.
how are the two cases different? in either f1(t) or f2(t), you have to try each case, ie, 1> t^2, t^2> 1 so both results, for f1(t) and f2(t) would be the same, no?
sorry -- typo. I meant
∫[-∞,∞] c*f1(t) dt = 1
Well, let's see. To determine if a function is a valid probability density function (PDF), we need a few conditions to be met:
1. The function must be non-negative for all values of t.
2. The integral of the function over the entire real line must equal 1.
Let's analyze f1(t) first. The square root of a negative number is imaginary, so for f1(t) to be non-negative, exp(-max(1, t^2)^2) must evaluate to a positive number for all t. However, for t > 1, max(1, t^2)^2 will always be greater than 1, meaning exp(-max(1, t^2)^2) will be less than 1, resulting in a negative number when subtracted from 12π.
So f1(t) cannot be a valid PDF. But fear not! We can multiply f1(t) by a positive constant c to satisfy condition 2 and make cf1(t) a valid PDF.
Now let's move on to f2(t). This function involves min(1, t^2), which will always be less than or equal to 1. Therefore, exp(-min(1, t^2)^2) will always be a positive number, regardless of the value of t. Moreover, 12π is a positive constant. So f2(t) is non-negative for all t.
To determine if f2(t) satisfies condition 2, we need to integrate it over the entire real line. Unfortunately, the integral of f2(t) is not analytically solvable, so it's hard to determine if it equals 1 without numerical methods. However, since f2(t) is always non-negative and involves exponential functions, it is likely that the integral will converge to a finite positive value. Therefore, it is reasonable to assume that f2(t) is a valid PDF.
In conclusion:
- f1(t) is not a valid PDF, but there exists a constant c such that cf1(t) is a valid PDF.
- f2(t) is likely a valid PDF, but further analysis or numerical methods are needed to confirm that its integral equals 1.
Hope that helps! Keep those probability functions rolling!
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 be non-negative for all values of the random variable.
2. The integral of the function over the entire range of the random variable must equal 1.
Let's analyze each function separately:
For f1(t) = 12π−−√exp(−max(1,t2)2), we can see that the function involves the maximum of 1 and t^2. Since the maximum function is non-negative for all real values of t, we only need to check the second condition.
The integral of f1(t) over the entire range of t can be computed as follows:
∫[all real numbers] f1(t) dt = ∫[from -∞ to ∞] 12π−−√exp(−max(1,t^2)^2) dt
It is important to note that the function inside the exponential, max(1, t^2), is equal to 1 for t^2 ≤ 1 and equal to t^2 for t^2 > 1.
To integrate over this range, we need to split the integral:
∫[from -∞ to ∞] 12π−−√exp(−max(1,t^2)^2) dt
= ∫[from -∞ to -1] 12π−−√exp(−(1)^2) dt
+ ∫[from -1 to 1] 12π−−√exp(−(t^2)^2) dt
+ ∫[from 1 to ∞] 12π−−√exp(−(t^2)^2) dt
The first and third integrals can be computed as definite integrals and will give finite values.
Now, let's consider the second integral from -1 to 1. Since the function max(1, t^2) is equal to t^2 in this range, the integral becomes:
∫[from -1 to 1] 12π−−√exp(−(t^2)^2) dt
= ∫[from -1 to 1] 12π−−√exp(−t^4) dt
This integral is a difficult one to solve analytically, but it is known that it is finite. Therefore, the entire integral over the real line is also finite.
Since the function f1(t) is non-negative for all values of t and its integral over the entire range is finite, f1(t) is a valid PDF for a continuous random variable.
Therefore, the answer to the first question is: Yes, it is a valid PDF.
Now let's move on to the second function, f2(t) = 12π−−√exp(−min(1,t^2)^2).
Similarly, we need to check if f2(t) satisfies both conditions for a valid PDF.
The function involves the minimum of 1 and t^2. The minimum function is non-negative for all real values of t, so we only need to check the second condition.
The integral of f2(t) over the entire range of t can be computed as follows:
∫[all real numbers] f2(t) dt = ∫[from -∞ to ∞] 12π−−√exp(−min(1,t^2)^2) dt
Again, we split the integral:
∫[from -∞ to ∞] 12π−−√exp(−min(1,t^2)^2) dt
= ∫[from -∞ to -1] 12π−−√exp(−(t^2)^2) dt
+ ∫[from -1 to 1] 12π−−√exp(−(1)^2) dt
+ ∫[from 1 to ∞] 12π−−√exp(−(1)^2) dt
The second integral from -1 to 1 becomes:
∫[from -1 to 1] 12π−−√exp(−(1)^2) dt
= ∫[from -1 to 1] 12π−−√exp(−1) dt
This integral is finite, but it is different from the first and third integrals, which involve a constant term.
Therefore, the integral of f2(t) over the entire range is not equal to 1.
Hence, the function f2(t) is not a valid PDF for a continuous random variable.
To determine if there is a constant c>0 that makes cf2 a valid PDF, we need to find a value of c that satisfies:
∫[all real numbers] cf2(t) dt = 1
This means finding a constant c such that the integral of cf2(t) over the entire range equals 1.
By multiplying f2(t) by c, we can see that the value of c does not affect the range over which the integrals are split. Therefore, the integral over the entire range will still be different from 1, regardless of the value of c.
Therefore, there is no constant c>0 that makes cf2 a valid PDF.
Hence, the answer to the second question is: No, it is not a valid PDF, and there is no constant c making cf2 a valid PDF.
In summary:
1. The function f1(t) is a valid PDF for a continuous random variable.
2. The function f2(t) is not a valid PDF for a continuous random variable, and there is no constant c making cf2(t) a valid PDF.
Therefore, the correct answer choices are: No, it is not a valid PDF, and there is no constant c making cf1 a valid PDF (for the first question), and No, it is not a valid PDF, and there is no constant c making cf2 a valid PDF (for the second question).