1

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!

Paul1911
  • 13
  • 3
  • 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 Jul 11 '21 at 21:16
  • @Dave Aah ok, that makes sense. The 3.4 as a result could consequently make sense, right? – Paul1911 Jul 11 '21 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 Jul 11 '21 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 $0$. – Glen_b Jul 11 '21 at 22:37
  • That seemed to be the implication, but then do we have a discrete-continuous mixture with $0.01$ probability of a $0$? – Glen_b Jul 11 '21 at 22:40
  • @Dave I checked it multiple times online, calculationwise, it is correct. If it is now also correct logically, I am a happy student. thanks! – Paul1911 Jul 12 '21 at 07:55
  • @Glen_b You're right, I forgot to mention that the support is $[0,\infty)$. – Paul1911 Jul 12 '21 at 07:55
  • Note that your $f$ is not $0$ at $x=0$ (since $1-0.28-0.71 = 0.01$), Since it therefore has a discrete component, you should be very careful when talking about $f'$. [It would also help later readers if you used the common conventions of using capital letters for cdfs and lower case for pdfs] – Glen_b Jul 13 '21 at 01:03
  • @Glen_b thank you, I adjusted it. – Paul1911 Jul 22 '21 at 18:47
  • The expectation is the integral (from $0$ to $\infty$) of $1-F.$ That's an elementary calculation: you can write the answer down by inspection. – whuber Jul 22 '21 at 19:45

1 Answers1

0

  • 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.

Subrata Pal
  • 81
  • 5
  • 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 $[0,\infty)$. – Paul1911 Jul 12 '21 at 07:59
  • In this mixture, *one of the weights is negative.* Such things arise among [sums of Gamma variables](https://stats.stackexchange.com/a/72486/919), 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 Jul 12 '21 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*. – Subrata Pal Jul 12 '21 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. – Subrata Pal Jul 12 '21 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. – Subrata Pal Jul 12 '21 at 15:28
  • A much **simpler way** to calculate the expectation without bothering about all these would be: $\int_0^\infty P(X>x)dx=\int_0^\infty (1-F(x))dx=E[X]$ The simpler proof is [here](https://stats.stackexchange.com/a/13377/327787), but the proof is exactly not applicable for your problem as it assumes the existence of a density. A rigorous proof is [here](https://math.stackexchange.com/a/64199/949249) – Subrata Pal Jul 12 '21 at 15:34
  • @SubrataPal This is really helpful, thank you! – Paul1911 Jul 13 '21 at 19:06
  • @Paul1911, You're most welcome... – Subrata Pal Jul 13 '21 at 19:43