# Solved – Find the joint distribution of two independent random variables

Two random variables such as \$X_{1}, X_{2},…,X_{n}\$ be iid's has pdf \$theta x^{theta-1}\$ where \$0<x<1\$ and \$Y_{1}, Y_{2},…,Y_{n}\$ be iid discrete random variables have power series distribution \$p(Y=y)=frac{gamma(y)theta^y}{c(theta)}\$ where \$y=0,1,2,…\$. Assume \$X\$'s and \$Y\$'s are independent.

I am trying to find the distribution of \$Z_{i}=X_{i}+Y_{i}\$.

Since \$X\$'s and \$Y\$'s are independent I can find the distribution of \$f(x,y)\$. Later, I can use the transformation to find the distribution of \$f(z,x)\$. Now I need integrate by \$Y\$ in order to find the marginal density of \$Z\$.

My question is how to find the limit for \$Y\$. Since I am dealing with continuous and discrete random variables.

Contents

Find it directly–avoid the middleman!

### Finding the distribution of $$Z_i$$

Because almost surely $$0 lt X_i lt 1$$ and $$Y_i$$ is one of the natural numbers $${0,1,2,ldots,},$$ consider any real number $$z ge 0$$ and write it as

$$z = y(z) + x(z)$$

where $$y(z) = lfloor z rfloor$$ is the greatest integer less than or equal to $$z$$ and $$x(z) = z – y(z)$$ is the fractional part left over. From these formulas we can reconstruct $$X_i$$ and $$Y_i$$ from $$Z_i$$ as

$$y(Z_i) = y(Y_i + X_i) = Y_i$$

and

$$x(Z_i) = x(Y_i + X_i) = Z_i – Y_i = X_i.$$

Thus, because $$Y_i$$ and $$X_i$$ are independent,

eqalign{F_{Z_i}(z) = Pr(Z_i le z) &= Pr(Y_i lt y(z)text{ or } (Y_i = y(z) text{ and } X_i le x(z))) \ &= Pr(Y_i lt y(z)) + Pr(Y_i = y(z))Pr(X_ile x(z)) \ &= F_{Y_i}(y(z)-1) + Pr(Y_i=y(z)) x(z)^theta. }

This is an effective formula for the distribution $$F_{Z_i}$$ of $$Z_i,$$ thereby answering the question. I will demonstrate its use by (a) computing its density and (b) integrating the density.

### Computing the density of $$Z_i$$

When $$z$$ is not an integer $$F_{Z_i}$$ is a differentiable function of $$z$$ with constant derivative $$1$$ because $$y$$ is differentiable (it's locally constant with derivative zero) and so, therefore, is $$x$$ because

$$frac{d}{dz} x(z) = frac{d}{dz}(z – y(z)) = 1 – 0 = 1.$$

Moreover, the summation does not change except when its upper endpoint $$y(z)$$ changes, which occurs only at the natural numbers. Still assuming $$z$$ is not a natural number, we compute the density of $$Z$$ simply by differentiating via the sum rule, product rule, and the chain rule:

$$f_{Z_i}(z) = frac{d}{dz}Pr(Z_i le z) = theta x(z)^{theta-1}Pr(Y_i = y(z)).$$

We may arbitrarily define $$f_{Z_i}$$ at the natural numbers: give it any finite values you like there. And, since $$Z_ige 0,$$ $$f_{Z_i}(z) = 0$$ for all $$zlt 0.$$ That completes the determination of the density.

This figure depicts the graph of $$f_{Z_i}$$ where $$Y_i$$ has a Poisson$$(3)$$ distribution and $$theta = 4.$$ The heights of the spikes in the graph are determined by the Poisson probabilities $$Pr(Y_i=y(z)),$$ while the shapes of the graph between the spikes are given by the density of $$X_i$$ (as scaled by $$Pr(Y_i=y(z))$$ and translated by $$y(z)$$).

### Integrating the density

As a check, let's verify that $$f_{Z_i}$$ is normalized to unit probability by integrating it, which we may do by breaking the integral into a sum of areas over the intervals $$[i, i+1)$$ for $$i=0, 1, 2, ldots:$$

eqalign{ int_mathbb{R} f_{Z_i}(z) dz &= int_0^infty theta x(z)^{theta-1}Pr(Y_i = y(z)) dz \ &= sum_{i=0}^infty int_{i}^{i+1} theta x(i+t)^{theta-1}Pr(Y_i = y(i+t)) d(i+t) \ &= sum_{i=0}^infty int_0^1 theta t^{theta-1} Pr(Y_i = i) dt \ &= sum_{i=0}^infty Pr(Y_i=i) = 1. }

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