The property I want is that I can change the variance with the mean fixed.

I know it is easy to come up one in continuous case, i.e normal distribution with mean $\mu\ $ and variance $\sigma^2$.

In discrete case, I am thinking negative binomial distribution is such one. However, I am not totally sure. Because negative binomial distribution $NB(r,p)\ $ has mean $\frac{pr}{1-p}\ $ and variance $\frac{pr}{(1-p)^2}=\text{mean}\cdot\frac{1}{1-p}\ $. It seems that if I change $p$ with mean fixed, the variance has been changed. However, if I changed $p$, mean would also change, right?

Can you give some such discrete probability distribution? And if negative binomial is one, why is that?

  • $\begingroup$ Should the discrete random variable be non-negative ? If this is not a requirement, Skellam distribution with parameters $m_1$ and $m_2$ has mean $m_1-m_2$ and variance $m_1+m_2$ $\endgroup$ – Sasha Oct 27 '11 at 6:19
  • $\begingroup$ $NB(3,1/4)$ and $NB(1,1/2)$ both have mean $1$, but their variances are $4/3$ and $2$, respectively. $\endgroup$ – Brian M. Scott Oct 27 '11 at 6:22
  • $\begingroup$ Your $NB(r,p)$ example seems allright. Assume you want a positive mean $m$ and a variance $v>m$, then choose $p=(v-m)/v$ and $r=m^2/(v-m)$. Hence for a fixed $m$ and a variable $v$, both $p$ and $r$ depend on $v$. $\endgroup$ – Did Oct 27 '11 at 8:04
  • $\begingroup$ In the negative binomial case, do you want the distribution with the same mean and the changed variance still to be a negative binomial distribution, or will you be satisfied with some other distribution? Do you want it still to be supported on nonnegative integers, or will you be satisfied with a distibution that takes non-integer values? $\endgroup$ – Michael Hardy Oct 27 '11 at 20:11

Here is a very cheap example. Let $X_a$ be the random variable that takes on the values $a$ and $-a$, each with probability $1/2$, where $a$ is a parameter. The mean stays fixed at $0$ and the variance varies. If you would like to change the mean to a fixed quantity other than $0$, the example can be easily adjusted.

Using the same idea, we take a random walk, say on the line, moving $1$ step left or $1$ step right, each with probability $1/2$. Let the random variable $W_n$ denote our net displacement (positive or negative) after $n$ steps.

We can do the same trick using the difference $U-V$ of two independent identically distributed binomials. As we vary the parameters $p$ and $n$, the mean stays at $0$. However, by adjusting the parameters, we can obtain arbitrary variance.

The following is a more important class of examples. Repeat an experiment independently $n$ times, with probability of success each time equal to $p$. Let the random variable $Y_{p,n}$ be the sample mean. Then $E(Y_{p,n})=p$ and $\text{Var}(Y_{p,n})=\frac{p(1-p)}{n}$. So we can decrease the variance by increasing the parameter $n$, a very useful fact.

The negative binomial family, specially if we look at the more general one with $r$ a positive real, can be made to have the right property by changing the named parameters. For example, use $\mu$ and $p$, where $\mu$ is the mean. Almost by definition, you can vary $p$ while keeping $\mu$ fixed, and thereby change the variance. But admittedly this is more than a little artificial.

  • $\begingroup$ In the example, $U$ and $V$ could even be arbitrary independent random variables with the same mean. Then $U-V$ has mean zero and variance the sum of the variances of $U$ and $V$. $\endgroup$ – Rasmus Oct 27 '11 at 7:25

In the negative binomial case, suppose $p=1/2$ and $r=5$, so the mean is $5$.

Now change $p$ to $1/4$ and let $r$ be $15$. Then the mean is still $5$, but the variance is different.

And it's still a negative binomial distribution.


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.