Chapter 6: Q.6.6 (page 275)

If X and Y are jointly continuous with joint density function fX,Y(x, y), show that X + Y is continuous with density function $f X + Y^{(t)} = q - q f X, Y (x, t - x) d x$

Short Answer

Expert verified

Find the CDF of X + Y first, and then use the theorem about the derivation of function where the argument is in the boundary to obtain the required PDF.

Step by step solution

Content Introduction

The derivative of the CDF is the probability density function f(x), abbreviated PDF if it exists. A distribution function FX describes each random variable X. (x).

Content Explanation

Lets find the CDF of Z:= X + Y firstly. Take any z. We have that

$F_{z} (z) = P (Z \leq z) = P (X + Y \leq z) = \int_{- \infty}^{\infty} \int_{- \infty}^{x - z} f_{X, Y} (x, y) d x d y$

Using the theorem from analysis about the derivation of function where the argument is in the boundary of integral, we have that

$f_{z} (z) = \frac{d}{d z} F_{z} (z) = \int_{- \infty}^{\infty} F_{X, Y} (x, x - z) d x$

which had to be proved.

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If X and Y are jointly continuous with joint density function fX,Y(x, y), show that X + Y is continuous with density function $f X + Y^{(t)} = q - q f X, Y (x, t - x) d x$

Short Answer

Step by step solution

Content Introduction

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If X and Y are jointly continuous with joint density function fX,Y(x, y), show that X + Y is continuous with density function fX+Y(t)=q−qfX,Y(x,t−x)dx

Short Answer

Step by step solution

Content Introduction

Content Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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If X and Y are jointly continuous with joint density function fX,Y(x, y), show that X + Y is continuous with density function $f X + Y^{(t)} = q - q f X, Y (x, t - x) d x$