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Questions:

  1. Can we talk about:
    variance of a deterministic variable?;
    covariance between a deterministic variable and a stochastic variable?;
    covariance between two deterministic variables?
  2. Are these concepts well defined in sample?; in population?

Motivation
Take a simple regression

$$y = \beta_0 + \beta_1 x + \varepsilon.$$

Suppose the regressor $x$ is stochastic. The OLS estimate of $\beta_1$ will be

$$\hat{\beta}_1=\frac{\widehat{\text{Cov}}(x,y)}{\widehat{Var}(x)}$$

where hats denote sample counterparts of the population concepts. No problem here.

Now suppose $x$ is deterministic. I am not sure if I can use terms like variance and covariance in this context. Should I exchange $\hat{\beta}_1=\frac{\widehat{\text{Cov}}(x,y)}{\widehat{Var}(x)}$ for something like

$$\hat{\beta}_1=\frac{\frac{1}{n-1}\sum(x_i-\bar{x})(y_i-\bar{y})}{\frac{1}{n-1}\sum(x_i-\bar{x})^2}$$

to be correct? But then again, how meaningful is $\bar{x}$ when $x$ is deterministic? So should I go all the way to

$$\hat{\beta}_1=\frac{\frac{1}{n-1}\sum_{i=1}^n(x_i-\frac{1}{n}\sum_{j=1}^n x_j)(y_i-\frac{1}{n}\sum_{j=1}^n y_j)}{\frac{1}{n-1}\sum(x_i-\frac{1}{n}\sum_{j=1}^n x_j)^2}?$$

I am picking on details here and this may not be too important; my main questions are listed at the top of the post.

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2 Answers 2

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All five questions have "yes" answers--but we have to be careful about what they mean.

  1. "Variance of a deterministic variable."

    Let's understand a "deterministic variable" to be a univariate dataset. It's just a bunch of values $X=x_1, x_2, \ldots, x_n$, with no probability model. By definition its variance is

    $$\text{Var}(X) = \frac{1}{n}\sum_{i=1}^n \left(x_i - \bar X\right)^2$$

    where $$\bar X = \frac{1}{n}\sum_{i=1}^n x_i$$ is its mean. There is no justification whatsoever to use $n-1$ instead of $n$ in any of these fractions--and this is never legitimately done--because no estimates are being made.

    We may always think of $X$ as defining a "population." This is the definition of a population variance.

  2. "Covariance between a deterministic variable and a stochastic variable."

    One way to understand this is to assume it refers to a sequence of the form $(x_1, Y_1), (x_2,Y_2), \ldots, (x_n,Y_n)$ where the $x_i$ are numbers and the $Y_i$ are random variables. Then we may define the random variable $$\bar Y = \frac{1}{n}\sum_{i=1}^n Y_i,$$ via which the covariance of $x$ and $Y$ is defined as

    $$\text{Cov}(x,Y) = \frac{1}{n}\sum_{i=1}^n (x_i - \bar x)(Y_i - \bar Y).$$

    It is a linear combination of the $Y_i$ and consequently is itself a random variable. This notation is frequently used as a shorthand in linear regression calculations.

  3. "Covariance between two deterministic variables."

    "Two deterministic variables" can be considered a dataset of ordered pairs $(x_1, y_1), (x_2,y_2), \ldots, (x_n,y_n)$. The covariance can be defined exactly as in (2) and interpreted similarly. In fact, this is a direct consequence of (1): after all, covariances are variances.

  4. "Are these concepts well defined in samples?"

    Because they are well-defined for any dataset, they are well-defined for a sample. Note that similar expressions with $n-1$ in the (outer) denominator are estimators: they are not the sample variance or sample covariance.

  5. "Are these concepts well defined in populations?"

    Because they are well-defined for any dataset, and a population can be considered a dataset (when fully enumerated), they are well-defined for a population.

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  • $\begingroup$ Can you please explain how we would use these definitions to calculate the correlation between $X$ and $Y$ when $(X_i, Y_i)=(Z_1+i, Z_2+i)$ with $Z_1, Z_2 \overset{\mathrm{i.i.d.}}{\sim}\mathcal{N}(0, 1)$ for deterministic $i=1,\ldots,n$? $\endgroup$
    – statmerkur
    Commented Jan 20, 2022 at 0:07
  • $\begingroup$ @Statmerkur The bilinearity of covariance implies the addition of $i$ does not change the covariance, while the independence of the $Z_i$ implies that covariance must either be $0$ or undefined. Because the $Z_i$ have finite variances, the covariance will be $0.$ $\endgroup$
    – whuber
    Commented Jan 20, 2022 at 4:46
  • $\begingroup$ Thank you. However, if I understand you correctly, the correlation estimate in this little R simulation n <- 100; cor(rnorm(n) + 1:n, rnorm(n) + 1:n) should be close to $0$ -- but it is actually close to $1$. The latter does make sense to me as the two sequences are identical up to a small noise component. Am I getting something wrong here? $\endgroup$
    – statmerkur
    Commented Jan 20, 2022 at 15:09
  • $\begingroup$ @stat You don't understand correctly if you expect that correlation to be near $0,$ but I cannot guess what the nature of the misunderstanding might be. $\endgroup$
    – whuber
    Commented Jan 21, 2022 at 14:44
  • $\begingroup$ I don't expect that correlation to be near $0$. Rather, I'm interested in how one would calculate the expected correlation in such a case. I was then confused about what you said above since, I thought, by the same reasoning the correlation between $X$ and $Y$ when $(X_i, Y_i)= (Z_{1,i}+i,Z_{2,i}+i)$ with $Z_{1,i}, Z_{2,i}\overset{\mathrm{i.i.d.}}{\sim}\mathcal{N}(0,1);i=1,\ldots,100$ would also have to be $0$. And I think the R code calculates the correlation in one particular realization of this sequence. $\endgroup$
    – statmerkur
    Commented Jan 21, 2022 at 15:53
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The simple answer to your first three questions is no: it makes no sense in general to talk about variance or covariance involving a deterministic variable.

However, if we begin with a deterministic variable but then use some method of imposing a probability distribution on it, it then becomes a random variable, and the concept of variance then makes sense. For example, any deterministic variable can become a random variable simply by imposing a degenerate (i.e., constant) distribution on it; in this case, the variance (along with its covariance with any other random variable) becomes 0.

A more interesting way of imposing a distribution on deterministic random variables is to use the empirical distribution, based on an observed sample. That is, if you observe $x_1,\dots,x_n$ in a sample, then we can define a discrete probability distribution on $x$ by $$P(x=x_0)=\frac1n \cdot\text{the number of $i$ such that $x_i=x_0$}$$ for all $x_0$. For example, in the case where $x_1,\dots,x_n$ are all distinct, we get $P(x=x_i)=\frac1n$ for each $i=1,\dots,n$. If we use this probability distribution on $x$, then the mean of $x$ becomes simply the sample mean $\overline x=\frac1n\sum_{i=1}^n x_i$, and the variance of $x$ becomes $\hat\sigma^2=\frac1n\sum_{i=1}^n (x_i-\overline x)^2$.

This same idea can be applied in the situation where we have observed vectors $(x_1,y_1),\dots,(x_n,y_n)$. We can define a joint probability distribution on $(x,y)$ again by using the empirical distribution, and the covariance between $x$ and $y$ then becomes $$\text{Cov}(x,y)=\frac1n\sum_{i=1}^n (x_i-\overline x)(y_i-\overline y)$$

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