Question

4.4-14. Let X and Y be random variables of the continu- ous type having the joint pdf f(x,y) = 8xy, 0<xsys 1. Draw a graph that illustrates the domain of this pdf.

Answer 1

(a)

The marginal pdf of X is

fx(x) = [ f(x, y)dy = | 8rydy = [4.ry?]) = 4x – 4x2

So,

fx() = 4.1 – 4.r”,0 <r <1

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The marginal pdf of Y is

0 0/ ght = [1,] = fiphias ) = xp(* *x)} ] = (6)^! f

So,

fy(y) = 4y3,0 <y<1

(b)

The mean of X is :

sex = E(X) – ["252(e)ko - 5' 12 – do'ko - Lua -

The mean of Y is :

$\mu_{Y}=E(Y)=\int_{0}^{1}yf_{Y}(y)dy=\int_{0}^{1}4y^{4}dy=\left [ \frac{4y^{5}}{5} \right ]_{0}^{1}=0.8$

EXY) – [ . uste: vyáyds = [ { $+ Páyite = b Mat JE

$=\int_{0}^{1}\left [\frac{8x^{2}}{3}-\frac{8x^{5}}{3} \right ]dx=\left [\frac{8x^{3}}{9}-\frac{8x^{6}}{18} \right ]_{0}^{1}=\frac{4}{9}=0.4444$

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$Cov(X,Y)=E(XY)-E(X)E(Y)=\frac{4}{9}-\frac{8}{15}\cdot 0.8=0.0178$

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Finding variances:

Date - ,-" ) = oplewiz" ) = (x)a

The variance of X is :

21 64 11 ož = Var(X) = E(X2) – [E(X))? = - 3 -225 = 11

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: - LA ) - 10/18 ) - punissa Y = 64,312

The variance of Y is

o = Var(Y) = E(Y2) – [E(X)]? = - 0.82 = 7

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The correlation coefficient is

p=- Cou(X,Y) Var(X)Var(Y) = 0.4930

(c)

The least square regression line is

» » » + rv .– 08+ 0.100. ME (-Ė) - 08+ y = Hy + p. (2 - x) = 0.8+ 0.4930 V2/75 V11/225 = 0.8 + 0.0243 (15.– 8)

Hence,

y = 0.8 +0.0243 (15. – 8)

Following is the graph:

Regression line (1,1)

As the graph shows that line goes approximately from the middle of triangle so it seems to be a good fit. That is line make sense.