I've been stuck on this for a while and it's been driving me crazy. Any help would be greatly appreciated.
I am trying to find the Expected Value of the following Probability Density Functions (where E is Euler's number) for the x values shown below. For some reason I get very different values from my calculations of the expected value when applying the discrete and continuous method of calculating expected value to my data set. If you can, please give me the expected value and the expected value function, I can compare it to mine.
First PDF function: 0.348*(E^((3488.034*x)-335.553)) for 0.1<=x<=1
Second PDF function: ((19.366*x)-0.697)^-5.111 for 0.1<=x<=1
Thank you, and, once again, I greatly appreciate the help.
What you have are not probability distributions over that support.
$$\displaystyle \int_{0.1}^1 f(x)\operatorname d x = \int_{0.1}^1 0.348\, e^{3488.034\, x-335.553}\operatorname d x \neq 1$$
So perhaps you need to either check the support, or normalise:
$$\mathsf E(X\mid X\in [0.1,1]) = \dfrac{\int_{0.1}^1 x f(x) \operatorname d x}{\int_{0.1}^1 f(x)\operatorname d x} \approx 0.9997133\ldots $$