1
$\begingroup$

Let's say I want to find the $n$-fold convolution of $n$ iid binomial random variables. My question is regarding the probability of success $p$. Would $p$ be the same for each random variable in the sample or is the correct way to allow $p$ to vary, that is, probability of success for each $X_i$ is $p_i$ for all $i = 1,\dotsc,n$. If it's the latter, then I don't understand the concept of iid. If these random variables are iid, then the expected value (at least) will be the same for each $X_i$; but if $p_i$'s are not necessarily equal, how can it be claimed that the expected values are equal?

$\endgroup$
  • 1
    $\begingroup$ If they're identically distributed, they have the same distribution, just like it sounds, so yes, the $p_i$ are all the same. $\endgroup$ – saulspatz Feb 10 '18 at 2:53
  • $\begingroup$ The second two letters of iid stand for 'identically distributed` so $p$ must be the same for all and $E(X_i)$ are all the same. $\endgroup$ – BruceET Feb 10 '18 at 7:52
1
$\begingroup$

Let's do it for $n=2.$ If, independently, $X_1, X_2 \sim \mathsf{Binom}(5, 1/3),$ then $Y = X_1 + X_2 \sim \mathsf{Binom}(10, 1/3).$ Intuitively, you can think of $X_1$ as the sum of the first five Bernoulli trials and $X_2$ as the sum of the second five. Perhaps the easiest formal proof is via moment generating functions. (The discussion extends to general $n$ without difficulty.)

The figure below was made by generating 100,000 realizations of $X_1, X_2,$ and $Y = X_1 + X_2,$ then making a histogram of the 100,000 sums $Y.$ Centers of open circles show the exact distribution of $\mathsf{Binom}(10, 1/3).$

enter image description here


R code for the sampling and the figure is shown below.

x1 = rbinom(10^5, 5, 1/3)
x2 = rbinom(10^5, 5, 1/3)
y = x1 + x2
hist(y, prob=T,br=(-1:10)+.5, col="skyblue2", main="Y ~ BINOM(10,1/3)")
  yy = 0:10; pdf=dbinom(yy, 10, 1/3)
  points(yy,pdf,col="red")
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Quick question before I accept the answer: When discussing convolutions, which is what $Y$ is, it’s implied that the random variables in our sample are iid, right? If so, then I’m confused by author of “Statistical Computing with R” when she writes “To simulate a convolution: Generate $x_1 ~ N(0,1)$ and $x_2 \sim N(3,1)$. Deliver $s=x_1+x_2$.” Clearly, $x_1$ and $x_2$ are not iid, so I don’t see how $s$ satisfies the definition of convolution. However, I do see $s$ as a more generalized form of a convolution where the distributions are permitted to be different. $\endgroup$ – Alana Feb 10 '18 at 16:11
  • $\begingroup$ Convolution is a general term. The most common case is for sums of iid RVs, but iid is not part of the definition. $\endgroup$ – BruceET Feb 10 '18 at 17:10
  • $\begingroup$ So at the very least the definition of convolution assumes independence amongst the rv’s? $\endgroup$ – Alana Feb 10 '18 at 17:25
  • 1
    $\begingroup$ I have only seen convolution used for independent random variables. Unfortunately, it is a term that gets used in more contexts than it gets rigorously defined. Without casting aspersions on the specific text you are using, I would not be surprised if the term has essentially 'colloquial' status in a book more concerned with computation than probability theory. And some 'data mining' books feel free to use terminology that is never defined. $\endgroup$ – BruceET Feb 10 '18 at 17:31
  • $\begingroup$ I see what you’re saying. It’s just frustrating reading the definition of a convolution in terms of iid rvs, and then providing an example of a convolution where the rvs are not identically distributed. Thank you for answering my questions. $\endgroup$ – Alana Feb 10 '18 at 17:34
1
$\begingroup$

There's no such thing as the convolution of two random variables. Two distributions (ie probability measures on $\Bbb R$) have a convolution, but a distribution is not a random variable.

The notion of the convolution of two probability distributions has a lot to do with independence, nothing necessarily to do with "identically distributed": If $X$ and $Y$ are independent random variables then the distribution of $X+Y$ is the convolution of the distribution of $X$ with the distribution of $Y$. That holds regardless of whether $X$ and $Y$ are identically distributed (and it doesn't say anything about the non-existent "convolution" of $X$ and $Y$.)

| cite | improve this answer | |
$\endgroup$

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.