# 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!

• The “exponential distribution” means something specific, not just that $e$ is raised to a power. You should not expect that nice $1/\lambda$ expected value.
– Dave
Commented Jul 11, 2021 at 21:16
• @Dave Aah ok, that makes sense. The 3.4 as a result could consequently make sense, right? Commented Jul 11, 2021 at 22:08
• I didn’t do any of the calculus to verify your work, but nothing screams out as a mistake. // There are many online resources for doing calculations such as these. Know how to do it by hand, since you need that skill on your exams, but it can be comforting to check a solution.
– Dave
Commented Jul 11, 2021 at 22:16
• To be a valid cdf we need to know what the support of $Y$ is, and what values of $y$ that the cdf applies for, and what values the cf takes otherwise. Please clarify. Note in particular that the cdf at $0$ is not $0$. Commented Jul 11, 2021 at 22:37
• That seemed to be the implication, but then do we have a discrete-continuous mixture with $0.01$ probability of a $0$? Commented Jul 11, 2021 at 22:40

## 1 Answer

• 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.

• Hi Subrata, thank you first of all. Looking at your last paragraph, I am not quite sure of the implications for me. The numbers were not a typo, they indeed do not sum up to 1. But I am not sure, what this now means for me to get a correct PDF/CDF - is it only the factor of 0.01 which must be added? Also, you are correct, I forgot the range. It is $[0,\infty)$. Commented Jul 12, 2021 at 7:59
• In this mixture, one of the weights is negative. Such things arise among sums of Gamma variables, for instance. But discussing that would take us far from the intent of this question, which is a basic exercise in integration only: the answer equals $\int_0^\infty f(x)\mathrm{d}x.$ It's unnecessary (and extra work) even to compute the pdf.
– whuber
Commented Jul 12, 2021 at 13:24
• Hi @Paul1911, it means that your cdf is a mixture of two exponential distributions with parameters 0.5 and 0.25 respectively, and one degenerate discrete random variable(X=0) with weights 0.28, 0.71, 0.01. This distribution does not have a probability density or a probability mass function in the usual sense as it is a mixture of both continuous and discrete random variables. So simply there is no pdf/pmf - but it has a cdf. Commented Jul 12, 2021 at 15:23
• More technically, the absolutely continuous distributions are dominated by the Lebesgue measure and the discrete distributions are dominated by the counting measures. This can neither be dominated by counting measure, nor by the Lebesgue measure, but a mixture of the two. Commented Jul 12, 2021 at 15:25
• And the expectation would be: $0.28/0.5+0.71/0.25+0.01*0=3.4$, which is the same as yours because the 3rd random variable is degenerate at 0. Commented Jul 12, 2021 at 15:28