Probability Theory – Problem 41: Compute , where is a real number and is a nonnegative integer;\\

The random variable $Y$ follows an exponential distribution with parameter $\lambda$. Let $[y]$ denote the greatest integer not exceeding $y$, and define the random variables $X = Y - [Y],\ Z = [Y]$.\\ (1) Compute $P(X \le x, Z = n)$, where $x$ is a real number and $n$ is a nonnegative integer;\\ (2) Determine the distributions of $X$ and $Z$;\\ (3) Discuss the independence of $X$ and $Z$.

1. Identifying the ranges of the variables and the event structure

Since $Y$ has an exponential distribution with parameter $\lambda$, we have $$f_{Y}(y)=\lambda e^{-\lambda y},\quad y>0.$$

Let $[Y]$ denote the floor function (the greatest integer not exceeding $Y$). Then $$Z=[Y],\quad X=Y-[Y].$$

Observe that: $Z$ takes nonnegative integer values $n=0,1,2,\dots$; for each sample point, $0\le X<1$, and $Y=Z+X$.

2. Rewriting the event $\{X\le x,Z=n\}$ in terms of $Y$

The event $Z=n$ means $$Z=n\Longleftrightarrow n\le Y<n+1.$$

When $Z=n$, we have $X=Y-n$, so $$X\le x,\ Z=n \Longleftrightarrow \begin{cases} n\le Y<n+1,\\ Y-n\le x \end{cases} \Longleftrightarrow \begin{cases} n\le Y<n+1,\\ Y\le n+x. \end{cases}$$

Therefore $$\{X\le x,Z=n\} =\{n\le Y\le \min(n+1,n+x)\},$$ and we proceed by cases according to the value of $x$.

3. Case analysis by the range of $x$

1. Case 1: $x\le 0$

Here $X\le x\le 0$ is required, but $X\ge 0$ always holds, so $$P(X\le x,Z=n)=0,\quad x\le 0.$$

2. Case 2: $0<x<1$

In this case $n+x<n+1$, so the integration interval is $[n,n+x]$, and $$P(X\le x,Z=n) =P(n\le Y\le n+x) =\int_{n}^{n+x}\lambda e^{-\lambda y}\,dy.$$

Evaluating the integral: $$\int_{n}^{n+x}\lambda e^{-\lambda y}\,dy =\left[-e^{-\lambda y}\right]_{y=n}^{y=n+x} =e^{-\lambda n}-e^{-\lambda(n+x)} =e^{-\lambda n}\big(1-e^{-\lambda x}\big).$$

Hence $$P(X\le x,Z=n) =e^{-\lambda n}\big(1-e^{-\lambda x}\big),\quad 0<x<1.$$

3. Case 3: $x\ge 1$

Since $X\in[0,1)$, when $x\ge 1$ the condition $X\le x$ is automatically satisfied, so $$\{X\le x,Z=n\}=\{Z=n\}.$$

Therefore $$P(X\le x,Z=n)=P(Z=n) =P(n\le Y<n+1) =\int_{n}^{n+1}\lambda e^{-\lambda y}\,dy.$$

Computing this gives $$\int_{n}^{n+1}\lambda e^{-\lambda y}\,dy =e^{-\lambda n}-e^{-\lambda(n+1)} =e^{-\lambda n}\big(1-e^{-\lambda}\big).$$

Hence $$P(X\le x,Z=n) =e^{-\lambda n}\big(1-e^{-\lambda}\big),\quad x\ge 1.$$

4. Summary (final answer for Part (1), stated piecewise)

For any real number $x$ and nonnegative integer $n$, $$P(X\le x,Z=n) = \begin{cases} 0,& x\le 0,\\[4pt] e^{-\lambda n}\big(1-e^{-\lambda x}\big),& 0<x<1,\\[4pt] e^{-\lambda n}\big(1-e^{-\lambda}\big),& x\ge 1. \end{cases}$$

1. Finding the distribution of $Z$

From the computation above, $$P(Z=n)=P(n\le Y<n+1) =\int_{n}^{n+1}\lambda e^{-\lambda y}\,dy =e^{-\lambda n}\big(1-e^{-\lambda}\big),\quad n=0,1,2,\dots.$$

Thus the probability mass function of $Z$ is $$P(Z=n)=\big(1-e^{-\lambda}\big)e^{-\lambda n},\quad n=0,1,2,\dots.$$

This is a geometric distribution starting at $0$ with parameter $1-e^{-\lambda}$.

2. Finding the distribution function of $X$

For $x<0$, since $X\ge 0$, $$P(X\le x)=0,\quad x<0.$$

For $0\le x<1$, by the law of total probability, $$P(X\le x) =\sum_{n=0}^{\infty}P(X\le x,Z=n).$$

Using the formula from Part (1) (for $0<x<1$; the value at $x=0$ is also consistent), $$P(X\le x,Z=n) =e^{-\lambda n}\big(1-e^{-\lambda x}\big),$$ we obtain $$P(X\le x) =\sum_{n=0}^{\infty}e^{-\lambda n}\big(1-e^{-\lambda x}\big) =\big(1-e^{-\lambda x}\big)\sum_{n=0}^{\infty}e^{-\lambda n}.$$

Since the geometric series satisfies $$\sum_{n=0}^{\infty}e^{-\lambda n} =\dfrac{1}{1-e^{-\lambda}},$$ it follows that $$P(X\le x) =\dfrac{1-e^{-\lambda x}}{1-e^{-\lambda}},\quad 0\le x<1.$$

For $x\ge 1$, since $X<1$ holds almost surely, $$P(X\le x)=1,\quad x\ge 1.$$

In summary, the distribution function of $X$ is $$F_{X}(x)=P(X\le x) = \begin{cases} 0,& x<0,\\[4pt] \dfrac{1-e^{-\lambda x}}{1-e^{-\lambda}},& 0\le x<1,\\[8pt] 1,& x\ge 1. \end{cases}$$

3. Finding the density function of $X$

For $0<x<1$, $F_{X}$ is differentiable, and the density is $$f_{X}(x)=\dfrac{d}{dx}F_{X}(x) =\dfrac{\lambda e^{-\lambda x}}{1-e^{-\lambda}},\quad 0<x<1.$$

For all other $x$, the density is $0$: $$f_{X}(x) = \begin{cases} \dfrac{\lambda e^{-\lambda x}}{1-e^{-\lambda}},& 0<x<1,\\[4pt] 0,& \text{otherwise}. \end{cases}$$

This shows that $X$ follows a truncated and renormalized exponential distribution on $[0,1)$.

4. Summary of Part (2)

$Z$ is a discrete random variable with distribution $$P(Z=n)=\big(1-e^{-\lambda}\big)e^{-\lambda n},\quad n=0,1,2,\dots.$$

$X$ is a continuous random variable supported on $[0,1)$ with density $$f_{X}(x)=\dfrac{\lambda e^{-\lambda x}}{1-e^{-\lambda}},\quad 0<x<1.$$

1. Computing $P(X\le x)P(Z=n)$

We know that $$P(Z=n)=e^{-\lambda n}\big(1-e^{-\lambda}\big).$$

We compare by cases in $x$:

If $x\le 0$, then $P(X\le x)=0$, so $$P(X\le x)P(Z=n)=0.$$

If $0<x<1$, then $$P(X\le x)=\dfrac{1-e^{-\lambda x}}{1-e^{-\lambda}},$$ so $$P(X\le x)P(Z=n) =\dfrac{1-e^{-\lambda x}}{1-e^{-\lambda}} \times e^{-\lambda n}(1-e^{-\lambda}) =e^{-\lambda n}\big(1-e^{-\lambda x}\big).$$

If $x\ge 1$, then $P(X\le x)=1$, so $$P(X\le x)P(Z=n) =P(Z=n) =e^{-\lambda n}\big(1-e^{-\lambda}\big).$$

2. Comparison with $P(X\le x,Z=n)$ from Part (1)

From Part (1), $$P(X\le x,Z=n) = \begin{cases} 0,& x\le 0,\\[4pt] e^{-\lambda n}\big(1-e^{-\lambda x}\big),& 0<x<1,\\[4pt] e^{-\lambda n}\big(1-e^{-\lambda}\big),& x\ge 1. \end{cases}$$

By comparison, for every real number $x$ and every nonnegative integer $n$, $$P(X\le x,Z=n)=P(X\le x)P(Z=n).$$

3. Independence conclusion and intuitive explanation

The above identity shows that the joint distribution of $X$ and $Z$ factors completely into the product of their marginal distributions. Therefore $$X \text{ and } Z \text{ are independent}.$$

1. For any real number $x$ and nonnegative integer $n$, $$P(X\le x,Z=n) = \begin{cases} 0,& x\le 0,\\[4pt] e^{-\lambda n}\big(1-e^{-\lambda x}\big),& 0<x<1,\\[4pt] e^{-\lambda n}\big(1-e^{-\lambda}\big),& x\ge 1. \end{cases}$$

2. Distribution of $Z$: $$P(Z=n)=\big(1-e^{-\lambda}\big)e^{-\lambda n},\quad n=0,1,2,\dots.$$

3. Distribution of $X$: $$f_{X}(x) = \begin{cases} \dfrac{\lambda e^{-\lambda x}}{1-e^{-\lambda}},& 0<x<1,\\[4pt] 0,& \text{otherwise}. \end{cases}$$

4. $X$ and $Z$ are independent.