2
$\begingroup$

This will seem like a very simple question to many of you; but I cannot understand part of the solution, so to give context I have had to unfortunately resort to typing the whole thing out, apologies.

Start of Question:

$100$ independent measurements (i.e. samples) are made of a random variable, which has an exponential distribution $\lambda e^{−\lambda x}$, and their average is found. Consider the probability distribution of this average. The expectation value of this distribution will be $\lambda^{-1}$.

What is the standard deviation of the distribution?

End of Question.


Start of Solution:

The sum of the $100$ values of $x$ is $$X=\sum_{i=0}^{i=N}x_i$$ and has an expectation value $$E(x)=\sum_{i=0}^{i=N}E(x_i)=N\cdot E(x)$$

Since $E(x) \propto x$, the expectation value of the average $\overline{x}=\dfrac{X}{N}$ is $$E\Big(\overline{x}\Big)=\dfrac{N\cdot E(x)}{N}=E(x)=\color{blue}{\dfrac{1}{\lambda}}$$ $\color{blue}{\fbox{$\text{as stated in the question}$}}$. Similarly, for the Variance $$Var(x)=\sum_{i=1}^{i=N}Var(x_i)=N\cdot Var(x)$$ Since $Var(x)\propto x^2$, the variance of the average $\overline{x}=\dfrac{X}{N}$ is $$Var\Big(\overline{x}\Big)=\dfrac{N\cdot Var(x)}{N^2}=\dfrac{1}{N}\cdot \color{red}{Var(x)}=\dfrac{1}{100}\cdot \color{red}{\dfrac{1}{\lambda^2}}\tag{?}\label{}$$ The Standard Deviation is therefore $$\sqrt{Var\Big(\overline{x}\Big)}=\dfrac{1}{10\lambda}$$

End of Solution.


I understand why $E(x)=\color{blue}{\dfrac{1}{\lambda}}$ as it was stated in the question as shown by the $\color{blue}{\mathrm{blue}}$ box. But what I don't understand is why $\color{red}{Var(x)}\stackrel{\eqref{*}}=\color{red}{\dfrac{1}{\lambda^2}}$. Could someone please explain to me how this was worked out?

$\endgroup$
  • 1
    $\begingroup$ A single random variable $Y$ with density $ae^{-ay}\mathbf 1_{y>0}$ has mean and variance $E(Y)=1/a$ and var$(Y)=1/a^2$. End of story. $\endgroup$ – Did Nov 18 '15 at 9:41
  • 1
    $\begingroup$ @Did That's so impressive, how do you do it? I feel like you deserve some rep for this, so why don't you make this as an answer and i'll accept it. $\endgroup$ – BLAZE Nov 18 '15 at 9:46
  • 1
    $\begingroup$ Hopefully somebody else will post this, or variants of this, as an answer, which, if you fully understand it, you will be able to accept, and everything shall be fine. More interesting to me would be to understand how in the first place you got stuck in this problem ('cause, seriously, everything in my o-so-impressive comment you already knew beforehand, no?). $\endgroup$ – Did Nov 18 '15 at 9:51
  • 1
    $\begingroup$ And now it seems your problem was to actually compute the variance, not to understand how the formula is even relevant to the "sample" situation in your question. In other words, you lost me and I would not be able to tell what your question really asks... $\endgroup$ – Did Nov 18 '15 at 9:53
  • $\begingroup$ @Did That's a very good point; the answer is I knew it once I saw it, but was unable to recall it by myself, as I keep saying probability and distribution theory is my weak spot. But thank you very much. $\endgroup$ – BLAZE Nov 18 '15 at 9:55
2
$\begingroup$

We have

$$E(X^2)=\int_0^\infty x^2 \lambda e^{-\lambda x}dx=\frac{2}{\lambda^2}$$

So,

$$Var(X)=E(X^2)-E(X)^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}$$

$\endgroup$
  • 1
    $\begingroup$ Thank you very much, as you can probably tell; distribution theory is not my strong point at all, very good answer $\endgroup$ – BLAZE Nov 18 '15 at 9:50
  • 1
    $\begingroup$ Note that the formular I mentioned is often useful to calculate variances. $\endgroup$ – Peter Nov 18 '15 at 9:50
  • $\begingroup$ Superb answer, thanks again $\endgroup$ – BLAZE Nov 18 '15 at 9:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.