I'm doing statistics homework and am stuck on a problem using integration.
The problem gives a distribution where for x>1, f(x) = k x^-6. I am then asked to "Determine the value of k for which f(x) is a legitimate pdf. "
To be a legitimate pdf, the integral of f(x)dx has to be equal to 1 (from -infinity to infinity).
I can do the integral, 1/(-5x^5) or k/(-5x^5), but how do I set this equal to 1 and solve for k without knowing x?
The only thing I have been able to come up with is that x = -.2^(1/5), or -.72477, which would make k = 1, thereby not changing the integral at all..
Hmmm.
1= INt (1->inf)k/x^6 dx=
= -k/5 * 1/x^5 (limits x=1+ to inf)
1= -k/inf + k/5
k= 5
check my thinkig
I have no clue!!!
Thanks bob! That's what I ended up working out and it turned out to be the right answer :)
To determine the value of k for which f(x) is a legitimate probability density function (pdf), you need to find the value that makes the integral of f(x) equal to 1.
In this case, the function is defined as f(x) = kx^(-6) for x > 1.
To proceed, you need to find the limits of integration. Since the question mentions integrating from -infinity to infinity, you can use those limits. However, since the distribution is defined only for x > 1, the integral will actually be from 1 to infinity.
To solve for k, set up the integral:
∫[1, ∞] kx^(-6) dx = 1
To evaluate the integral, you can integrate the function term by term:
∫[1, ∞] kx^(-6) dx = k ∫[1, ∞] x^(-6) dx
Integrating x^(-6), we get:
k ∫[1, ∞] x^(-6) dx = k * [(x^(-5)) / -5] [1, ∞]
Now, let's calculate the integral:
= k * [(∞^(-5)) / -5 - (1^(-5)) / -5]
As x goes to infinity (∞), x^(-5) becomes 0, and since 1^(-5) is equal to 1, we have:
= k * [0 - (1 / -5)]
Simplifying further:
= k * (1 / 5)
Setting the result equal to 1:
k * (1 / 5) = 1
To solve for k, multiply both sides of the equation by 5:
k = 5
Therefore, the value of k for which f(x) is a legitimate pdf is k = 5.