# What is the distribution of a random variable that is the product of the n dependent random variables?

If random variables $$X$$, $$Y$$ are dependent, and $$X>0$$ $$Y>0$$, we have $$P(XY\leq z)=P(X\leq z/Y)=\int_{0}^{\infty} \int_{0}^{z/y} f_{XY}(x,y) dx dy$$.

How about the case of $$n$$ dependent random variables? i.e., how to use the joint PDF $$f_{X_1,X_2,\ldots,X_n}(x_1,x_2,\ldots,x_n)$$ to obtain the CDF $$P(X_1 X_2\cdots X_n\leq z)$$?

I have an idea, but not sure if it is right.

$$P(X_1 X_2\cdots X_n\leq z)\\=E(P(X_1 \leq \frac{z}{X_2\cdots X_n}\mid X_2=x_2,\ldots,X_n=x_n))\\=\int_0^\infty \cdots \int_0^\infty dx_2\cdots dx_n \int_0^{\frac{z}{x_2\cdots x_n}}f_{X_1,X_2,\ldots,X_n}(x_1,x_2,\ldots,x_n)dx_1$$.

• Do you have a particular multivariate family in mind or just any multivariate distribution? – JimB Apr 29 at 22:01
• For any multivariate distribution. I have an idea, but not sure if it is right. – Xinlei Yu Apr 29 at 22:14

## 1 Answer

Let $$X_1,....,X_n$$ be random variables with joint density function $$f(x_1,...,x_n)$$.

Consider the transformation $$g(x_1,...,x_n)=(x_1...x_n,x_2,...,x_n)$$. The inverse transformation is $$g^{-1}(z_1,...,z_n)=(\frac{z_1}{z_2...z_n},z_2,...,z_n)$$ and it follows that the jacobian of the inverse transformation is $$|J|=\frac{1}{z_2...z_n}$$.

Then by the transformation theorem the density function of the vector $$(z_1,...,z_n)$$ is $$f_Z(z_1,...,z_n)=f(\frac{z_1}{z_2...z_n},z_2,...,z_n)|\frac{1}{z_2...z_n}|$$ and the marginal density function of $$Z_1$$ can be written as:

$$f_{Z_1}(z_1)=\int_0^\infty...\int_0^\infty f(\frac{z_1}{z_2...z_n},z_2,...,z_n)|\frac{1}{z_2...z_n}| dz_2...dz_n$$

Note that the solution above can be obtained by applying the Leibniz rule to the solution you proposed. Take $$F(\frac{z}{x_2...x_n},...,x_n)-F(0,x_2,...,x_n)=\int_0^\frac{z}{x_2,...,x_n} f_{X_1,...,X_n}(x_1,...,x_n) dx_1$$ and try to calculate $$\frac{d}{dz}\int_0^\infty...\int_0^\infty F(\frac{z}{x_2...x_n},...,x_n)-F(0,x_2,...,x_n) dx_2...dx_n.$$