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.

Thanks in advance.

  • 1
    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
up vote 2 down vote accepted

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.

Figure: Graph of the PDF

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 $i=0, 1, 2, \ldots:$

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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.