9 Let the probability density of the two-dimensional randomvariable (ξ, η) be(4e**, x > 0;f;(x)=•(0,6e", y>0;-6yf,(v) =;0,x<0.y<0.FindE(2 { +3 n ).

Given: fξx=4e-4x,x>00 , x≤0 , fηx=6e-6y,y>00 , y≤0 we know that Ex=∫xfxdx So,…

Answered: 9 Let the probability density of the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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9 Let the probability density of the two-dimensional random
variable (ξ, η) be
(4e**, x > 0;
f;(x)=•
(0,
6e", y>0;
-6y
f,(v) =;
0,
x<0.
y<0.
Find
E(2 { +3 n ).

Expert Solution

Step 1

Given:

$f_{ξ} (x) = \{\begin{cases} 4 e^{- 4 x}, x > 0 \\ 0, x \leq 0 \end{cases}, f_{η} (x) = \{\begin{cases} 6 e^{- 6 y}, y > 0 \\ 0, y \leq 0 \end{cases}$

we know that $E (x) = \int x f (x) d x$

So,

$E (ξ) = \int_{0}^{\infty} x f_{ξ} (x) d x = 4 \int_{0}^{\infty} x e^{- 4 x} d x = 4 {[- \frac{x e^{- 4 x}}{4} - \frac{e^{- 4 x}}{16}]}_{0}^{\infty} = \frac{1}{4}$

$E (η) = \int_{0}^{\infty} y f_{η} (y) d y = 6 \int_{0}^{\infty} y e^{- 6 y} d y = 6 {[- \frac{y e^{- 6 y}}{6} - \frac{e^{- 6 y}}{36}]}_{0}^{\infty} = \frac{1}{6}$

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