Show that $g_k$ is the probability density of $X_k$

Question

Let $X_k$ ($1 \leqslant k \leqslant n$) be real valued continuous random variables with joint probability distribution function $f_{X_1,\cdots,X_n}$ given by $$f_{X_1,\cdots,X_n}(x_1, \cdots ,x_n) = g_1(x_1) \cdots g_n(x_n)$$ where $g_1, \cdots, g_n$ are some functions.

Does anyone know how to show $g_k$ is the probability density of $X_k$ ($1 \leqslant k \leqslant n$)?

I am getting confused when it said $f_{X_1,\cdots,X_n}$ is the joint probability distribution function because we always assume $f_{X_1,\cdots,X_n}$ as density function.

Answer 1

As usual we can find the marginal density of $X_1$ by integrating out the remaining $n-1$ variables. So \begin{align*} f_{X_1}(x_1) &= \int_{-\infty}^{\infty} \dots \int_{-\infty}^\infty g_1(x_1)\dots g_n(x_n) \: dx_2\dots dx_n \\ &= g_1(x_1) \prod_{i=2}^n\int_{-\infty}^\infty g_i(x_i) \: dx_i \end{align*} And similar we get for $k=2,\dots,n$, that $$f_{X_k}(x_k) = g_k(x_k) \prod_{i\neq k} \int_{-\infty}^\infty g_i(x_i) \: dx_i.$$ If in addition we add the natural condition that $\int_{-\infty}^\infty g_i(x_i) \: dx_i = 1$ for $i=1,\dots,n$, then it can be seen that $f_{X_k}=g_k$.

Answer 2

Strictly speaking, this isn't true, and it's for a rather silly reason. Suppose $X,Y$ are independent. Then their joint density function $f_{XY}$ satisfies $f_{XY}(x, y) = f_{X}(x)f_{Y}(y)$. Now set $g_{X} = 1/2 f_{X}, g_{Y} = 2f_{Y}$. Then $g_{X}, g_{Y}$ satisfy your equation, but they aren't densities since they don't integrate to $1$.

However, we can show if $f_{XY}(x,y) = g_{X}(x)g_{Y}(y)$ for functions $g_X, g_Y$, then they are independent. Indeed, $$P(X \leq s, Y \leq t) = \int_{x \leq s} \int_{y \leq t} g_{X}(x)g_{Y}(y) \ dy dx = \left(\int_{x \leq s} g_{X}(x) \ dx\right)\left(\int_{y \leq t} g_{Y}(y) dy\right)$$

The above formula implies the following three equalities $$P(X \leq s) = \left(\int_{x \leq s} g_{X}(x) \ dx\right)\left(\int_{\mathbb{R}} g_{Y}(y) dy\right)$$ $$P(Y \leq t) = \left(\int_{\mathbb{R}} g_{X}(x) \ dx\right)\left(\int_{y \leq t} g_{Y}(y) dy\right)$$ $$1 = \left(\int_{\mathbb{R}} g_{X}(x) \ dx\right)\left(\int_{\mathbb{R}} g_{Y}(y) dy\right)$$

whence $P(X \leq s)P(Y \leq t) = \left(\int_{x \leq s} g_{X}(x) \ dx\right)\left(\int_{y \leq t} g_{Y}(y) dy\right) = P(X \leq s, Y \leq t)$ as desired.

This argument also implies that the density of $X$ is $x \mapsto g_{X}(x) \int_{\mathbb{R}} g_{Y}(y) dy$, and that of $Y$ is $y \mapsto g_{Y}(y) \int_{\mathbb{R}} g_{X}(x) dx$.

Extending this to $n$ random variables uses the same idea.