Expected Value of Exponential CDF I am given the following CDF and I want to calculate its expected value:
$F(Y \leq y) =1-( 0.28e^{-0.5y} + 0.71e^{-0.25y})$
Creating the PDF:
$f(Y \leq y) = \frac{71\mathrm{e}^{-\frac{x}{4}}+56\mathrm{e}^{-\frac{x}{2}}}{400}$
Now I have of course read that $E(y) = 1/\lambda$ - But I don't see a clear $\lambda$ here.
Using $\int_0^\infty f(Y \leq y)y~dy$ (following this video) returns $3.4$, if I did it correctly. Is this calculation applicable here and did I do it correctly?
Because, following the wikipedia article and its visualisations, I can see that $P(x = E(x)) = 0.5P(x = 0)$ for all $\lambda$ shown as an example. This is not the case for my result of 3.4.
Thank you already!
 A: *

*First note that, the cdf of of an exponential distribution with parameter $\lambda$ would be $F(x)=1-e^{-\lambda x}I_{x\in [0, \infty)}$.


*Now, if we have two random variables $X_1$ and $X_2$ with cdf respectively $F_1(x)$ and $F_2(x)$, the mixture distribution (with mixing proportion $\alpha$ and $1-\alpha$ resp.) would be:
$$\alpha F_1(x)+(1-\alpha)F_2(x)$$
with the corresponding mean $\alpha E(X_1) + (1-\alpha)E(X_2)$.


*Now, I'm going to answer a simpler version of your question. When the cdf is: $$1-(0.28e^{-0.5y}+0.72e^{-0.25y}) = 0.28(1-e^{-0.5y}) + 0.72(1-e^{-0.25y})$$
in the range of $y\in [0, \infty)$, you can get values of $\lambda$'s as $(0.5, 0.25)$ respectively.


*For your question, I guess that the cdf is:
$$F(Y\le y) = [1-(0.28e^{-0.5y}+0.71e^{-0.25y})]I_{y\in [0, \infty)}$$
You should have written the range of $y$ also.


*Now this expression can be rewritten as:
$$0.28(1-e^{-0.5y})I_{y\in [0, \infty)} + 0.71(1-e^{-0.25y})I_{y\in [0, \infty)} + 0.01I_{y\in [0, \infty)}$$
Basically, if my guess is correct, a small mass of weightage $0.01$ on the point $0$. So the density is not exactly what you have written, or possibly there is some typo where the weights are like 0.28 and 0.72, or 0.29 and 0.71.
