The question asked whether or not f(x) = 0.5e^x when x is less than or equal to zero and f(x) = 0.5e^(-x) when x is greater than or equal to zero is a valid probability density.
I know that it is the integral... but usually you have a domain that is not a single point. Should I increase the domain slightly??
huh? Looks to me like the domain is all real numbers. The fact that x=0 is included in both parts doesn't really matter, since e^0 = e^-0 = 1
So, since the functions are symmetric, all you need is
∫[0,∞] 0.5 e^-x dx = 1/2
OH MY GOODNEss! I never even thought of that! Thank you so much!
To determine whether the given function f(x) is a valid probability density, we need to check if it satisfies two conditions: non-negativity and integrability.
1. Non-negativity: For a valid probability density function, f(x) must be non-negative for all values of x in its domain.
In this case, when x is less than or equal to zero, f(x) = 0.5e^x, and when x is greater than or equal to zero, f(x) = 0.5e^(-x).
For x ≤ 0: Since e^x is always positive, multiplying it by 0.5 will still yield a positive value. Therefore, f(x) = 0.5e^x is non-negative when x ≤ 0.
For x ≥ 0: Since e^(-x) is always positive, multiplying it by 0.5 will still yield a positive value. Therefore, f(x) = 0.5e^(-x) is non-negative when x ≥ 0.
Since f(x) is non-negative for both x ≤ 0 and x ≥ 0, it satisfies the non-negativity condition.
2. Integrability: For a valid probability density function, the integral of f(x) over its entire domain must be equal to 1.
To check the integrability, we can integrate f(x) over its given domain. However, as you correctly mentioned, the given domain consists of a single point (x = 0), which is not commonly used. To make it easier to compute the integral, it is common practice to extend the domain slightly.
In this case, we can extend the domain slightly to include a small interval around x = 0. For example, we can extend the domain to the interval (-ε, ε), where ε is a small positive number.
So, the new function becomes:
f(x) = 0.5e^x, if x < -ε
f(x) = 0.5e^(-x), if x > ε
f(x) = c, if -ε ≤ x ≤ ε
Here, c is a constant that ensures f(x) remains continuous at x = ε and x = -ε.
Now, you can calculate the integral of this extended function over its extended domain (-ε, ε). If the integral evaluates to 1, then f(x) is a valid probability density function.
In summary, to determine if the given function f(x) = 0.5e^x when x is less than or equal to zero and f(x) = 0.5e^(-x) when x is greater than or equal to zero is a valid probability density, extend the domain slightly and calculate the integral of f(x) over the extended domain.