Suppose that $X_1,X_2,\ldots,X_n$ is a random sample from a distribution with probability function:

$$p_X(x)=\begin{cases} 1/2 &x=-1,1 \\ 0 & \text{otherwise} \end{cases}$$

Now if we define the average

$\bar{X}= \frac{1}{n} \sum X_i$, I need to show that that for an odd $n$ the probability function for $\bar{X}$ is:

$$p_{\bar{X}} (x)= \frac{\binom{n}{\frac{n}{2} (x+1)}}{2^n}$$ for $x=\pm 1/n,\pm 3/n,\ldots,\pm 1$, $0$ otherwise.

Since $P[\sum X_i=k] =P [\bar{X}=k/n ]$, I thought it would be easier to first derive the probability function for the sum. Since they are still $2^n$ n-tuples and each one is equilikely, I need to count the ones whose sum is $k$ for:

$k=\pm 1,\pm 3,\ldots \pm n-2,\pm n$ (since even numbers are not a feasible combination)

The problem here though is that I do not immediately recognize how I could count all these tuples. Could you please help me with this? If it is easier to proceed another way, I am of course all ears.

  • $\begingroup$ Could you elaborate on what you understand "$\binom{n}{n(x+1)/2}$" to be? I ask this because for many people it is defined to be the count you seek! $\endgroup$
    – whuber
    Jul 2, 2014 at 16:49
  • $\begingroup$ @whuber I cannot quite interpret $\frac{n(x+1)}{2}$ What is the intuition here? $\endgroup$
    – JohnK
    Jul 2, 2014 at 17:26
  • 2
    $\begingroup$ As $x$ ranges over $\pm 1/n, \pm 3/n,$ etc, $n(x+1)/2$--when sorted--ranges over $0, 1, \ldots, n$. It represents the number of $1$'s that are summed (along with the remaining $-1$s) to produce $nx$. $\endgroup$
    – whuber
    Jul 2, 2014 at 17:51
  • $\begingroup$ @whuber Okay, thank you. It's still a little blurry in my head but I'll get it eventually. $\endgroup$
    – JohnK
    Jul 2, 2014 at 18:01

1 Answer 1


If $Z \sim Bernoulli(\frac12)$, then $X=2Z-1$ has a Rademacher distribution where $X=-1$ or $1$ with equal probability. Let $X_1, ..., X_n$ denote indepedent Rademacher random variables.

Then, the pmf of:

$$ S = \sum_{i=1}^n X_i = (2 \sum_{i=1}^n Z_i) - n = 2 Y -n$$

where $Y \sim Binomial(n,\frac12)$ (since the sum of $n$ Bernoulli's is $Binomial(n,p)$).

That is: $Y \sim Binomial(n,\frac12)$ with pmf:

$$f(y) = 2^{-n} \binom{n}{y} \quad \quad \text{for } y = {0,1,\dots,n}$$

Then, the pmf of $\bar X = \frac{S}{n} = \frac1n(2 Y - n)$ can be obtained directly via the method of transformations, which yields:

$$2^{-n} \binom{n}{\frac{1}{2} n \left(\bar{x}+1\right)}$$

... which is the result you seek. Since $y = {0,1,\dots, n}$, the domain of support of $\bar X = \frac1n(2 Y - n)$ will be:

  • for odd-valued $ n$: $\quad \bar x= \pm \frac1n,\pm \frac3n,\ldots, \pm 1$
  • for even-valued $n$: $\quad \bar x= 0, \pm \frac2n,\pm \frac4n,\ldots,\pm 1$
  • 1
    $\begingroup$ +1. I did not know that distribution had a name, thank you. $\endgroup$
    – JohnK
    Jul 2, 2014 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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