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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?

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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 $m_1+m_2$ –  Sasha Oct 27 '11 at 6:19
    
$NB(3,1/4)$ and $NB(1,1/2)$ both have mean $1$, but their variances are $4/3$ and $2$, respectively. –  Brian M. Scott Oct 27 '11 at 6:22
    
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 $v$. –  Did Oct 27 '11 at 8:04
    
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? –  Michael Hardy Oct 27 '11 at 20:11

2 Answers 2

up vote 2 down vote accepted

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.

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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 $V$. –  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.

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