You are not logged in.
Pages: 1
Suppose X is a continuous random variable with density
f(x)=x^3, for 0<x<t and f(x) = 0 otherwise.
Compute the expectation of X.
I know that first you integrate x^3 on the interval from 0 to t, so this gives you
t^4 divided by 4.
But after this, I'm unsure of what to do.
First, you need to multiply your density function by the value it produces. In other words, you need to integrate x^4, not x^3.
I'm not that great at probability, so I'm using intuition from here on out. Wikipedia says that the expectation is the integral of x^4 from -infinity to infinity, but this produces an expected value of t^5 / 5, which is greater than t (the maximum value of x) for every t greater than the fourth root of 5. My guess is to set up 2 integrals equal to each other to find the middle value, like this:
From here we get the equation
Like I said, this is just a guess. It makes intuitive sense to me, but my intuition is usually wrong when it comes to probability.
Wrap it in bacon
Offline
First, you calculate the value of t. You know that the total probability must be 1, therefore
Hence the expected value is
Last edited by JaneFairfax (2007-11-01 12:53:48)
Offline
Pages: 1