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Chapter 3: Problem 52

Suppose that \(X\) and \(Y\) are independent random variables with probability density functions \(f_{X}\) and \(f_{Y}\). Determine a one-dimensional integral expression for \(P[X+\) \(Y

Short Answer

Expert verified
The one-dimensional integral expression for the probability \(P[X+Y<x]\) is: \(P[X+Y<x] = \int_{-\infty}^{x} f_X(t) \cdot f_Y(x-t) \, dt\)

Step by step solution

01

Recall the formula for the probability of the sum of two independent random variables

Since X and Y are independent random variables with probability density functions f_X and f_Y, we can express the probability P[X+Y<x] as the convolution of the two density functions. That is: \(P[X+Y<x] = (f_X * f_Y)(x)\), where '*' denotes convolution.
02

Write down the convolution integral

The convolution of two functions can be calculated using the integral: \((f_X * f_Y)(x) = \int_{-\infty}^{\infty} f_X(t) \cdot f_Y(x-t) \, dt\) So, \(P[X+Y<x] = \int_{-\infty}^{\infty} f_X(t) \cdot f_Y(x-t) \, dt\)
03

Express the probability in terms of a one-dimensional integral

We can rewrite the probability P[X+Y<x] as an integral from -鈭� to x instead of -鈭� to 鈭�, because the only time events contribute is when \(X+Y < x\): \(P[X+Y<x] = \int_{-\infty}^{x} f_X(t) \cdot f_Y(x-t) \, dt\) This gives us the desired one-dimensional integral expression for the probability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Random Variables
In probability theory, random variables are said to be independent if the occurrence of one variable does not influence the occurrence of another. Think of flipping two coins; whether the first coin lands on heads or tails does not affect the outcome of the second coin.For independent random variables \( X \) and \( Y \), their joint probability behaves in a special manner. Specifically, the probability of both events happening together is the product of their individual probabilities. This means we can write:
  • For any two events \( A \) and \( B \): \( P(X \in A \text{ and } Y \in B) = P(X \in A) \cdot P(Y \in B) \)
Understanding independence is crucial for working with the probability of sums of random variables. It allows us to use the convolution of their probability density functions to evaluate expressions like \( P[X+Y
Probability Density Function
A probability density function (PDF) is a mathematical function that describes the likelihood of a continuous random variable taking on a specific value. It helps us understand how probabilities are distributed over different outcomes. For instance, if you have a probability density function \( f_X(x) \) for a random variable \( X \), you can determine how likely different values of \( X \) are.Key properties of PDFs include:
  • The total area under the PDF curve is equal to 1; this represents the certainty of an occurrence (some event will definitely happen).
  • The probability that a random variable \( X \) falls within an interval \( [a, b] \) is given by the area under the PDF curve from \( a \) to \( b \), represented as \( \int_{a}^{b} f_X(x) \, dx \).
PDFs are essential when working with independent random variables. They lay the groundwork for the convolution process, which is needed to find the probability of sums of such variables.
Integral Expression for Probability
To find the probability that the sum of two independent random variables \( X \) and \( Y \) is less than a certain value \( x \), we use an integral expression. This method is facilitated by the concept of convolution, which combines the probability density functions of \( X \) and \( Y \).The integral representation of this probability is:\[ P[X+Y
  • \( f_X(t) \cdot f_Y(x-t) \): *Here, each value of \( t \) represents a potential value that \( X \) may take, and \( x-t \) signifies the corresponding value for \( Y \) so that their sum is less than \( x \).* The product \( f_X(t) \cdot f_Y(x-t) \) calculates how probable it is for both variables to take values \( t \) and \( x-t \) simultaneously.
  • \( \int_{-\infty}^{x} ... \, dt \): *This integral sums up these probabilities across all possible values of \( t \) that lead to \( X+YThis intuitive application of integrals to probability allows us to precisely calculate the likelihood of linear combinations of random variables.

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    Most popular questions from this chapter

    Each element in a sequence of binary data is either 1 with probability \(p\) or 0 with probability \(1-p .\) A maximal subsequence of consecutive values having identical outcomes is called a run. For instance, if the outcome sequence is \(1,1,0,1,1,1,0\), the first run is of length 2, the second is of length 1, and the third is of length \(3 .\) (a) Find the expected length of the first run. (b) Find the expected length of the second run.

    Let \(X_{1}, X_{2}, \ldots\) be independent continuous random variables with a common distribution function \(F\) and density \(f=F^{\prime}\), and for \(k \geqslant 1\) let $$ N_{k}=\min \left\\{n \geqslant k: X_{n}=k \text { th largest of } X_{1}, \ldots, X_{n}\right\\} $$ (a) Show that \(P\left\\{N_{k}=n\right\\}=\frac{k-1}{n(n-1)}, n \geqslant k\). (b) Argue that $$ f_{X_{N_{k}}}(x)=f(x)(\bar{F}(x))^{k-1} \sum_{i=0}^{\infty}\left(\begin{array}{c} i+k-2 \\ i \end{array}\right)(F(x))^{i} $$ (c) Prove the following identity: $$ a^{1-k}=\sum_{i=0}^{\infty}\left(\begin{array}{c} i+k-2 \\ i \end{array}\right)(1-a)^{i}, \quad 0

    A and B play a series of games with A winning each game with probability \(p\). The overall winner is the first player to have won two more games than the other. (a) Find the probability that \(\mathrm{A}\) is the overall winner. (b) Find the expected number of games played.

    Find the expected number of flips of a coin, which comes up heads with probability \(p\), that are necessary to obtain the pattern \(h, t, h, h, t, h, t, h\).

    If \(R_{i}\) denotes the random amount that is earned in period \(i\), then \(\sum_{i=1}^{\infty} \beta^{i-1} R_{i}\), where \(0<\beta<1\) is a specified constant, is called the total discounted reward with discount factor \(\beta .\) Let \(T\) be a geometric random variable with parameter \(1-\beta\) that is independent of the \(R_{i} .\) Show that the expected total discounted reward is equal to the expected total (undiscounted) reward earned by time \(T\). That is, show that $$ E\left[\sum_{i=1}^{\infty} \beta^{i-1} R_{i}\right]=E\left[\sum_{i=1}^{T} R_{i}\right] $$
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