# Which distribution has an expected product equal to one?

Consider $n$ values $x_{1}$ to $x_{n}$ such that $x_{1}x_{2}\cdots x_{n} = 1$. Importantly, these values are non-independent, as their product must be equal to 1. The trivial case of $x_{i}=1 \quad \forall i$ is of no interest.

I am trying to find which distribution could generate such a dataset. More precisely, for $n$ non-independent draws from a continuous random variable $X$, where $E(x_{1}x_{2}\cdots x_{n}) = 1$, can we say anything about the underlying distribution?

• Wait - do you want the product to always equal one, or only equal one in expectation? Your first paragraph says the former, your second says the latter. – Paul Jul 18 '16 at 10:21
• The way my dataset is constructed is such that the product of these values always equals to one. Hence, my question is to think which distribution could reproduce such data generation process. The latter has to be studied in expectation terms, as it relates to draws from a distribution – luchonacho Jul 18 '16 at 10:22
• If there must be a product = 1 constraint on all realizations of the random variable (rather than the expected value being 1), then assuming any distribution with a sum to zero constraint for $\log X$ (google e.g. for "normal distribution simplex") would do that. – Björn Jul 18 '16 at 10:25
• @luchonacho The observed draws from your distribution always equal one, so the data generation process should generate draws with expected product always equaling one. If you only require expected product one, you might never get any draw with product one. It could just be above one sometimes and below one other times, and average out to one. That's not matching your data - your data has product equal to one always, not above one sometimes and below one other times. – Paul Jul 18 '16 at 11:20
• The equality to one is an approximation. It is always equal to one up to a given approximation degree. – luchonacho Jul 18 '16 at 11:31

Let $X_1, X_2, \ldots, X_{n-1}$ be standard normal random variables (mean zero, variance one). Let $$X_n = - \sum_{i=1}^{n-1} X_i.$$ Note that $X_n$ is normally distributed with mean zero and variance $n-1$. We have $$\sum_{i=1}^n X_i = 0$$ Exponentiating both sides, we have $$\exp\left(\sum_{i=1}^n X_i\right) = \prod_{i=1}^n e^{X_i} = 1$$ Let $Y_i = e^{X_i}$, and we have $$\prod_{i=1}^n Y_i = 1$$ as requested. The variables $Y_i$ are log-normally distributed.
This is the simplest way to get your desired result, but there are other ways too. For example, you could subtract off $\tfrac{1}{n} \sum_{j=1}^n X_j$ from each $X_i$, which would yield identically distributed (but not independent) variables that sum to zero. More precisely, let $$X'_i = X_i - \frac{1}{n} \sum_{j=1}^n X_j$$ then we have $$\sum_{i=1}^n X'_i = \sum_{i=1}^n X_i - n\left(\frac{1}{n} \sum_{j=1}^n X_j\right) = 0$$
Exponentiating both sides and letting $Y_i = e^{X'_i}$, we would again get log-normally distributed variates whose product is always 1.
• This is a good approach. Looking at your argument, you should discover you need none of the assumptions--not normality, not standardization, not even existence of expectations. All you require is that $X_1, \ldots, X_{n-1}$ be any $n-1$-variate distribution whatsoever. By defining $X_n = -\sum_{i=1}^{n-1}X_i$, you guarantee that all $n$ of the variables sum to $0$--which is a constant, whence the product of their exponents is $1$. Conversely, this construction shows every such distribution $(Y_1,\ldots, Y_n)$ arises in this manner provided the $Y_i$ are positive almost surely. – whuber Jul 18 '16 at 14:37
• The problem of this answer is that distribution is not the same for each $x_{i}$. The variance of $x_{n}$ differs from the rest. The point of the question is to find a distribution $G(\cdot)$ such that the expected value of the multiplication of $n$ draws is equal to one. – luchonacho Jul 18 '16 at 15:05
• In the second construction in my answer, all the variables $Y_i$ have the same distribution, provided that all the $X_i$ have the same distribution. – Paul Jul 18 '16 at 15:19