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.
unanswered
Determine whether the function f_2 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_2 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_2 a valid PDF.
No, it is not a valid PDF, and there is no constant c making cf_2 a valid PDF.
None of the above
No, it is not a valid PDF, and there is no constant c making c f_2 a valid PDF.
To determine whether a function is a valid probability density function (PDF), we need to check if it satisfies the properties of a PDF.
For a function to be a valid PDF, it must satisfy the following conditions:
1. The function must be non-negative for all values of t.
2. The integral of the function over its entire domain must equal 1.
Let's analyze f_1(t):
f_1(t) = (1/√(2π)) * e^(-max(1, t^2)/2)
1. Non-negativity:
We can see that the exponential function e^(-max(1, t^2)/2) is always positive. Therefore, the overall function f_1(t) is non-negative for all values of t.
2. Integral:
To check the integral, we need to determine the domain of integration. Since t can take any value on the real line, the domain of integration would be from -∞ to +∞.
∫[from -∞ to +∞] f_1(t) dt
= ∫[from -∞ to +∞] (1/√(2π)) * e^(-max(1, t^2)/2) dt
We observe that the function f_1(t) is integrable over its entire domain since the exponential function decays rapidly. Therefore, we can calculate the integral.
However, analyzing the integral mathematically can be complicated. Therefore, we will evaluate the integral using numerical methods (e.g., numerical approximation, software).
To summarize:
The function f_1(t) is a valid PDF if the integral ∫[from -∞ to +∞] (1/√(2π)) * e^(-max(1, t^2)/2) dt equals 1.
Now let's analyze f_2(t) following the same steps.
f_2(t) = (1/√(2π)) * e^(-min(1, t^2)/2)
1. Non-negativity:
Similar to f_1(t), the exponential function e^(-min(1, t^2)/2) is always positive, so f_2(t) is non-negative for all values of t.
2. Integral:
Again, we need to evaluate the integral ∫[from -∞ to +∞] (1/√(2π)) * e^(-min(1, t^2)/2) dt.
Similarly to f_1(t), analyzing the integral analytically is complicated. Thus, we need to evaluate it numerically.
To summarize:
The function f_2(t) is a valid PDF if the integral ∫[from -∞ to +∞] (1/√(2π)) * e^(-min(1, t^2)/2) dt equals 1.