# Is an integral of a density function of a random vector a density or probability?

I think the expression $$\int_{\mathbb R} f_{X_1, X_2}(x,x) dx$$ is the density of the distribution of the RV $X_1 - X_2$ at 0. It is not the probability that $X_1 - X_2=0$.

It (the probability that $X_1 - X_2=0$) should be $P(X_1 = X_2) = \iint_D f_{X_1}(x) f_{X_2}(y)\ dx \ dy$, where $D = \{(x,y) \in {\mathbb R}^2: x = y\}$.

My question is:

For some integral of a density function of a random vector, how can one tell if the integral is the density of some r.v. at some value, or the probability of some other r.v. at some value?

Is it a density if and only if the dimension of the integral is strictly less than the dimension of the original random vector, and a probability if and only if the dimension of the integral is equal to the dimension of the original random vector?

Thanks and regards!

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Can you clarify what dimension (of an integral or a random vector) means? – Srivatsan Sep 13 '11 at 12:21
For dimension of an integral, two possiblities: the dimension of the subset the integral is taken, or the number of iterated integrals that the original integral can be converted to when possible. I think the two are the same thing? For dimension of a random vector, it is the dimension when viewing the r.v. as a vector. – Tim Sep 13 '11 at 12:43

I am sure others will provide a mathematically correct answer. I will instead focus on some intuitive ideas and heuristics.

Dimensional analysis. Imagine that the quantity $x$ is not a number but a physical quantity like length. In that case, the dimension of both $x$ and $dx$ are $[L]$ (which is the same as saying they have the unit $\text{m}$, for meters). Remember that the density has the dimension that is the inverse of this, i.e., $[L^{-1}]$. Why? Because $$\Pr[a \leq X \leq b] = \int_a^b f_X(x) dx,$$ and the probability is always a number. Analogously, if $X_1$ and $X_2$ are two random variables with dimension $[L]$ each, then the joint density $f_{X_1, X_2}(x_1, x_2)$ has the dimension $[L^{-2}]$. This is because: $$\Pr[(X_1, X_2) \in \mathcal D] = \int_D f_{X_1, X_2}(x_1,x_2) dA,$$ where $dA = dx_1 dx_2$ is an area element.

Now, coming to your integral, it has the dimension of $[L^{-2}] \cdot [L] = [L^{-1}]$. This reminds us of a one-dimensional density, and not a probability. Thus the density option does look more plausible.

Degenerate vs. non-degenerate events. I do not think I am using the proper terminology here; I will appreciate if someone helps improve the answer.

This is essentially an elaboration of Robert's answer (but in my own words). By degenerate, I essentially mean that the event is a subset of zero measure. Let's take the event $X_1 = X_2$. This is a line $l$ (through the origin, slope $1$) in the plane; so it has zero measure. If the joint distribution has a continuous density, then we will expect that the integral over any region $\mathcal D$ of zero will be zero. As an example of this, take the case when $X_1$ and $X_2$ are iid and have density $f_X(x)$.

Before I proceed further, note that in general the joint variable need not have such a density, however well-behaved $X_1$ and $X_2$ are. For example, take $X_1 \sim U(0,1)$, and define $X_2 := X_1$. In this case, the joint random variable takes values only on the line $l$. The above logic breaks down because the density now takes the form $\delta(x_1 - x_2)$, and is not continuous.

Going back to your example, we have convinced ourselves that the probability should be zero, but the integral you wrote down would not be in general. This suggests that the integral cannot represent the probability.

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@Tim Please let me know if something is not clear. – Srivatsan Sep 13 '11 at 13:17
Thanks! I understand your reply basically. As for dimensional analysis, to avoid guessing, what does the notation $[L]$ mean, some unit for measuring something? Is it used often in Physics? – Tim Sep 13 '11 at 13:25
@Tim You're welcome. By "my reply", do you mean this answer? (As for dimensional analysis, see en.wikipedia.org/wiki/Dimensional_analysis#Definition. It's essentially the same as units, but they use different symbols for reasons not clear to me. And yes, it's used more often in physics.) – Srivatsan Sep 13 '11 at 13:28
Yes, I mean I can understand this answer. – Tim Sep 13 '11 at 13:33