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We roll a 6-sided die a large number of times.
Calculating the difference (absolute value) between a roll and its preceding roll, are the differences expected to be uniformly distributed?
To illustrate with 10 rolls:
roll num result diff
1 1 0
2 2 1
3 1 1
4 3 2
5 3 0
6 5 2
7 1 4
8 6 5
9 4 2
10 4 0
Would the diff values be uniformly distributed?
No it is not uniform
You can count the $36$ equally likely possibilities for the absolute differences
second 1 2 3 4 5 6
first
1 0 1 2 3 4 5
2 1 0 1 2 3 4
3 2 1 0 1 2 3
4 3 2 1 0 1 2
5 4 3 2 1 0 1
6 5 4 3 2 1 0
which gives a probability distribution for the absolute differences of
0 6/36 1/6
1 10/36 5/18
2 8/36 2/9
3 6/36 1/6
4 4/36 1/9
5 2/36 1/18
Using only the most basic axioms about probabilities and real numbers, one can prove a much stronger statement:
The difference of any two independent, identically distributed nonconstant random values $X-Y$ never has a discrete uniform distribution.
(An analogous statement for continuous variables is proven at Uniform PDF of the difference of two r.v.)
The idea is that the chance $X-Y$ is an extreme value must be less than the chance that $X-Y$ is zero, because there is only one way to (say) maximize $X-Y$ whereas there are many ways to make the difference zero, because $X$ and $Y$ have the same distribution and therefore can equal each other. Here are the details.
First observe that the hypothetical two variables $X$ and $Y$ in question can each attain only a finite number $n$ of values with positive probability, because there will be at least $n$ distinct differences and a uniform distribution assigns them all equal probability. If $n$ is infinite, then so would be the number of possible differences having positive, equal probability, whence the sum of their chances would be infinite, which is impossible.
Next, since the number of differences is finite, there will be a largest among them. The largest difference can be achieved only when subtracting the smallest value of $Y$--let's call it $m$ and suppose it has probability $q = \Pr(Y=m)$--from the largest value of $X$--let's call that that one $M$ with $p = \Pr(X=M).$ Because $X$ and $Y$ are independent, the chance of this difference is the product of these chances,
$$\Pr(X-Y = M - m) = \Pr(X=M)\Pr(Y=m) = pq \gt 0.\tag{*}$$
Finally, because $X$ and $Y$ have the same distribution, there are many ways their differences can produce the value $0.$ Among these ways are the cases where $X=Y=m$ and $X=Y=M.$ Because this distribution is nonconstant, $m$ differs from $M.$ That shows those two cases are disjoint events and therefore they must contribute at least an amount $p^2 + q^2$ to the chance that $X-Y$ is zero; that is,
$$\Pr(X-Y=0) \ge \Pr(X=Y=m) + \Pr(X=Y=M) = p^2 + q^2.$$
Since squares of numbers are not negative, $0 \le (p-q)^2,$ whence we deduce from $(*)$ that
$$\Pr(X-Y=M-m)=pq \le pq + (p-q)^2 = p^2 + q^2 - pq \lt p^2 + q^2 \le \Pr(X-Y=0),$$
showing the distribution of $X-Y$ is not uniform, QED.
A similar analysis of the absolute differences $|X-Y|$ observes that because $X$ and $Y$ have the same distribution, $m=-M.$ This requires us to study $\Pr(X-Y=|M-m|) = 2pq.$ The same algebraic technique yields almost the same result, but there is the possibility that $2pq=2pq+(p-q)^2$ and $2pq+p^2+q^2=1.$ That system of equations has the unique solution $p=q=1/2$ corresponding to a fair coin (a "two-sided die"). Apart from this exception the result for the absolute differences is the same as that for the differences, and for the same underlying reasons already given: namely, the absolute differences of two iid random variables cannot be uniformly distributed whenever there are more than two distinct differences with positive probability.
Let's apply this result to the question, which asks about something a little more complex.
Model each independent roll of the die (which might be an unfair die) with a random variable $X_i,$ $i=1, 2, \ldots, n.$ The differences observed in these $n$ rolls are the numbers $\Delta X_i = X_{i+1}-X_i.$ We might wonder how uniformly distributed these $n-1$ numbers are. That's really a question about the statistical expectations: what is the expected number of $\Delta X_i$ that are equal to zero, for instance? What is the expected number of $\Delta X_i$ equal to $-1$? Etc., etc.
The problematic aspect of this question is that the $\Delta X_i$ are not independent: for instance, $\Delta X_1 = X_2-X_1$ and $\Delta X_2 = X_3 - X_2$ involve the same roll $X_2.$
However, this isn't really a difficulty. Since statistical expectation is additive and all differences have the same distribution, if we pick any possible value $k$ of the differences, the expected number of times the difference equals $k$ in the entire sequence of $n$ rolls is just $n-1$ times the expected number of times the difference equals $k$ in a single step of the process. That single-step expectation is $\Pr(\Delta X_i = k)$ (for any $i$). These expectations will be the same for all $k$ (that is, uniform) if and only if they are the same for a single $\Delta X_i.$ But we have seen that no $\Delta X_i$ has a uniform distribution, even when the die might be biased. Thus, even in this weaker sense of expected frequencies, the differences of the rolls are not uniform.
On an intuitive level, a random event can only be uniformly distributed if all of its outcomes are equally likely.
Is that so for the random event in question -- absolute difference between two dice rolls?
It suffices in this case to look at the extremes -- what are the biggest and smallest values this difference could take?
Obviously 0 is the smallest (we're looking at absolute differences and the rolls can be the same), and 5 is the biggest (6 vs 1).
We can show the event is non-uniform by showing that 0 is more (or less) likely to occur than 5.
At a glance, there are only two ways for 5 to occur -- if the first dice is 6 and the second 1, or vice versa. How many ways can 0 occur?
As presented by Henry, differences of uniformly distributed distributions are not uniformly distributed.
To illustrate this with simulated data, we can use a very simple R script:
barplot(table(sample(x=1:6, size=10000, replace=T)))
We see that this produces indeed a uniform distribution. Let's now have a look at the distribution of the absolute differences of two random samples from this distribution.
barplot(table(abs(sample(x=1:6, size=10000, replace=T) - sample(x=1:6, size=10000, replace=T))))
Others have worked the calculations, I will give you an answer that seems more intuitive to me. You want to study the sum of two unifrom r.v. (Z = X + (-Y)), the overall distribution is the (discrete) convolution product :
$$ P(Z=z) = \sum^{\infty}_{k=-\infty} P(X=k) P(-Y = z-k) $$
This sum is rather intuitive : the probability to get $z$, is the sum of the probabilities to get something with X (noted $k$ here) and the complement to $z$ with -Y.
From signal processing, we know how the convolution product behave:
You can understand what happen here : as $z$ move up (the vertical dotted line) the common domain of both rectangle move up then down, which correspond to the probability to get $z$.
More generally we know that the only functions that are stable by convolution are those of the gaussian familly. i.e. Only gaussian distribution are stable by addition (or more generally, linear combination). This is also meaning that you don't get a uniform distribution when combining uniform distributions.
As to why we get those results, the answer lies in the Fourrier decomposition of those functions. The Fourrier transformation of a convolution product being the simple product of the Fourrier transformations of each function. This give direct links between the fourrier coefficients of the rectangle and triangle functions.
If $x$ and $y$ are two consecutive dice rolls, you can visualize $|x-y| = k$ (for $k = 0, 1, 2, 3, 4, 5$) as follows where each color corresponds to a different value of $k$:
As you can easily see, the number of points for each color is not the same; therefore, the differences are not uniformly distributed.
Let $D_t$ denote the difference and $X$ the value of the roll, then $P(D_t = 5) = P(X_t = 6, X_{t-1} = 1) < P((X_t, X_{t-1}) \in \{(6, 3), (5, 2)\}) < P(D_t = 3)$
So the function $P(D_t = d)$ is not constant in $d$. This means that the distribution is not uniform.
