Assume that X and Y are jointly continuous random variables with joint probability density function given by f(x,y)=beg∈{cases}136(3x−xy+4y) if 0&lt;x&lt;2 and 1&lt;y&lt;3∖0 othrewiseend{cases} Find E[X] and E[Y].

Step 1 E[X]=μX =∫xxg(x)dx =∫02x(118(x+8))dx =118∫02(x2+8x)dx =118[x33+8x22]02 =118[83+8(2)] =818[73] =5654 =2827 Step 2 E[Y]: μY=∫yyh(y)dy =16y(16(y+1))dy =16(52+246) =16[16+246] =7636 =199

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Tazmin Horton

Answered question

2021-11-04

Assume that X and Y are jointly continuous random variables with joint probability density function given by
$f (x, y) = b e g \in {c a s e s} \frac{1}{36} (3 x - x y + 4 y) if 0 < x < 2 and 1 < y < 3 ∖ 0 o t h r e w i s e e n d {c a s e s}$
Find E[X] and E[Y].

Answer & Explanation

gotovub

Skilled2021-11-05Added 98 answers

Step 1

E [X] = μ_{X}

= \int_{x} x g (x) d x

= \int_{0}^{2} x (\frac{1}{18} (x + 8)) d x

= \frac{1}{18} \int_{0}^{2} (x^{2} + 8 x) d x

= \frac{1}{18} {[\frac{x^{3}}{3} + 8 \frac{x^{2}}{2}]}_{0}^{2}

= \frac{1}{18} [\frac{8}{3} + 8 (2)]

= \frac{8}{18} [\frac{7}{3}]

= \frac{56}{54}

= \frac{28}{27}

Step 2
E[Y]:

μ_{Y} = \int_{y} y h (y) d y

= \frac{1}{6} y (\frac{1}{6} (y + 1)) d y

= \frac{1}{6} (\frac{52 + 24}{6})

= \frac{1}{6} [\frac{16 + 24}{6}]

= \frac{76}{36}

= \frac{19}{9}

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