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Given two continuous random variables $X$ and $Y$, the joint cumulative distribution function $F_{X,Y}$ is defined as $$F_{X,Y}(x,y)=\mathbb{P}(X\le x, Y\le y)=\displaystyle\int_{-\infty}^{x}\int_{-\infty}^{y} f_{X,Y}(t_1,t_2)\mathrm{d}t_1\mathrm{d}t_2$$, where $f_{X,Y}$ is the joint probability density function of $X$ and $Y$.

The second partial derivative $\dfrac{\partial^2}{\partial x\partial y}F_{X,Y}(x,y)$ gives the joint probability density $f_{X,Y}(x,y)$.

But what does, say, partial derivatives $\dfrac{\partial}{\partial x}F_{X,Y}(x,y)=\int_{-\infty}^{y} f_{X,Y}(x,t_2)\mathrm{d}t_2$ and $\dfrac{\partial}{\partial y}F_{X,Y}(x,y)=\int_{-\infty}^{x} f_{X,Y}(t_1,y)\mathrm{d}t_1$ represent? Do they have any particular interpretation?

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The first-order partial derivatives of a multivariate joint distribution function can be considered as giving the density of the differentiated variable, jointly with the cumulative probability of the other variable(s). One simple way to see this interpretation is to convert the partial derivative to a density integral, integrated over the other dimensions. From the fundamental theorem of calculus we can write the partial derivative as:

$$\begin{aligned} \frac{\partial}{\partial x} F_{X,Y} (x, y) &= \int \limits_{- \infty}^y f_{X,Y} (x, t) dt \\[6pt] &= \int \limits_{- \infty}^y f_{Y|X} (t|x) f_X(x) dt \\[6pt] &= \int \limits_{- \infty}^y f_{Y|X} (t|x) dt \times f_X(x) \\[6pt] &= \mathbb{P} (Y \leqslant y | X = x) f_X (x). \end{aligned}$$

This shows that the partial derivative gives us the joint density over the line $Y \leqslant y, X = x$ (within the two-dimensional space of the two random variables). The partial derivative with respect to $y$ has an analogous interpretation.

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  • $\begingroup$ We could simplify the last term into $\mathbb{P}(Y \leq y, X = x)$ this also makes it more intuitive, at least to me $$\frac{\partial}{\partial x} \mathbb{P}(Y \leq y, X \leq x) = \mathbb{P}(Y \leq y, X = x)$$ $\endgroup$ – Sextus Empiricus Mar 2 '18 at 10:23
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    $\begingroup$ That statement seems to me to be false, since $\mathbb{P}(X=x) = 0$ for any continuous random variable $X$, and so the joint probability would also be zero. I think you have to keep the density of $X$ separate to the probability mass of $Y$. $\endgroup$ – Ben Mar 2 '18 at 20:48
  • $\begingroup$ That's right I should have used $$\mathbb{P}(Y \leq y, x-\frac{1}{2} dx \leq X \leq x+\frac{1}{2} dx)$$ $\endgroup$ – Sextus Empiricus Mar 3 '18 at 11:01
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    $\begingroup$ Still need to divide by $dx$, otherwise it's still zero. ;) $\endgroup$ – Ben Mar 3 '18 at 13:24
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    $\begingroup$ @user106860: I have added some additional steps to the working to show this explicitly. As you can see, it comes from splitting the joint density into the product of a marginal and conditional density. $\endgroup$ – Ben May 3 at 1:09
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If you take the joint CDF over xy and derive it over just one of the variables - you're left with marginal PDF for that same variable.

Let's prove using a simple joint distribution of two i.i.d. RVs X and Y ~Expo(1)

On one hand, we can get to the marginal PDF through the joint PDF: $$ F_{XY}(x,y) = \iint_{XY} f_{XY}(x,y) dxdy = \int_{0}^{\infty }\int_{0}^{\infty } e^{-x}e^{-y}dxdy \\ f_{XY}(x,y) = \frac{\partial^{2} }{\partial x \partial y} F_{XY}(x,y) = e^{-x}e^{-y} \\ f_{X}(x) = \int_{Y}f_{XY}(x,y)dy = e^{-x} \int_{0}^{\infty } e^{-y}dy = e^{-x} $$ Where last equation simplifies to e^(-x) because PDF of standalone y integrates to 1.

Alternatively we can go directly from joint CDF to marginal PDF: $$f_{X}(x) = \frac{\partial }{\partial x} F_{XY}(x,y) = e^{-x} \int_{0}^{\infty } e^{-y}dy = e^{-x}$$

Marginal PDF, if you're unfamiliar, is basically the PDF of X standalone, "freed up" from Y.

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  • $\begingroup$ I don't think this is true. Per the other answers, it is obvious that you still have the CDF of the variable that is NOT differentiated. This is inherently different from a marginal distribution. In a marginal distribution of X, we've integrated over all values of Y. Here, though - generally speaking - you still have Y <= <some value> which you have to pick. $\endgroup$ – user3814483 Apr 22 at 10:19

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