# 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.

• have you tried moment generating functions instead? the transformation theorem might be awkward because one rv is cts and one is discrete. – Taylor Oct 16 '17 at 3:19
• @Taylor, Thanks. I tried that one too. Since $f(x)$ is $Beta(\theta,1)$, it doesn't have explicit form of MGF. – score324 Oct 16 '17 at 3:58
• Well, a beta distribution do have an explicit mgf, it is an confluent hypergeometric function If that helps you is another question ... – kjetil b halvorsen Oct 16 '17 at 6:57
• This "power series distribution" is literally any distribution defined on the natural numbers. Specifically, the latter is determined by a sequence of probabilities $p_0,p_1,p_2,\ldots$ which sum to unity. Given any $\theta\gt 0$, then for each natural number $y$ simply define $\gamma(y)=\theta^{-y}p_y$. This makes your question awfully broad! Could you explain what you mean by "the limit for $Y$"? – whuber Oct 16 '17 at 14:40
• @ whuber, It should be an interval $[a, b]$ of the definite integral of the marginal density $f(z)$. – score324 Oct 16 '17 at 16:29

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_i\le 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 $$Z_i\ge 0,$$ $$f_{Z_i}(z) = 0$$ for all $$z\lt 0.$$ 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. }