I'm taking a calculus probability course and I'm stuck on my homework. The question is as follows:
We consider the exponential distribution on $\Omega = R≥0$ with parameter $λ = 4 $. Thus the probability density of $X$ is: $4e^{-4x}$ the expectation of $X$ is $E(X) = {1\overλ} = {1\over4}$ and its variance is $Var(X) = λ^{−2} = {1\over16}$. We define a stochastic variable $Y : Ω → R$ by $Y (x) = e^x$.
1) Calculate $P(Y ≤ a)$ for $a ∈ R$
2) Check that the expected value of $Y$ is $4\over3$
3) Calculate the variance and the standard deviation of $Y$
So there are multiple things I don't understand. The first one is how the $E(x)$ can be $1\over4$. if you say that: $Ω=R≥0$ how can there be an expected value? Since the set contains an infinite amount of numbers.(I know the formula so it has to be true but I'm still confused)
For the first question I'm not quite sure what to do. I know that the formula for the cumulative distribution is: $1-e^-4x$ so should I plug $a$ in for $x$ and than calculate it? or should I solve something like: $E(e^x) = \int_0^a 4*e^{-4e^x}$ The last one seems more likely but I don't think that's the correct formula.
For the second question I think I have to integrate from $0$ to $∞$ but I'm not sure which function I should integrate.
I think I can do the third question.
Any help would be appreciated!
That you can have a finite expected value depends of the "weight" described by the density of each number, although there are infinite ones.
To make it clear just consider the naturals $\Bbb N = \{1,2,3,\ldots\}$. There are obviously infinite of it but if I gave each natural $n \in \Bbb N$ a "weight" of e.g. $\frac{1}{2^n}$ the sum of all weights are finite, and the expected value becomes $$\sum_{n=1}^\infty n\cdot\frac{1}{2^n} = 1$$ So it's finite although you sum an infinite amount of numbers. But they are "weighted" by $\frac{1}{2^n}$ what makes each summand "small enough" to keep the series finite.
For the expected value use the formula for measurable functions $f$ $$E[f(X)] = \int_0^\infty \,f(x) \cdot 4e^{-4x}\,dx$$