Consider two continuous random variables X and Y with joint density function f(x,y)={x+y o≤x≤1,0≤y≤10 otherwise P(X&gt;0.8,Y&gt;0.8) is?

Step 1 Given information: The two continuous random variables X and Y has joint density function defined as: f(x,y)={x+y o≤x≤1,0≤y≤10 otherwise Then, P(X&gt;x,Y&gt;y)=∫y1∫x1f(x,y)dxdy Step 2 P(X&gt;0.8,Y&gt;0.8)=∫y=0.81∫x=0.81(x+y)dxdy =∫y=0.81{∫x=0.81(x+y)dx}dy =∫y=0.81[x22+xy]0.81dy =∫y=0.81{(12+y)−(0.822+0.8y)}dy =∫y=0.81{12+y−0.32−0.8}dy =∫y=0.81(0.18+0.2y)dy =[0.18y+0.2y22]0.81 =[(0.18+0.1)−(0.18×0.8+0.1×0.82)] =0.072 The required probability is 0.072.

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Jerold

Answered question

2021-06-03

Consider two continuous random variables X and Y with joint density function
$f (x, y) = {\begin{cases} x + y o \leq x \leq 1, 0 \leq y \leq 1 \\ 0 o t h e r w i s e \end{cases}$
$P (X > 0.8, Y > 0.8)$ is?

Answer & Explanation

pattererX

Skilled2021-06-04Added 95 answers

Step 1
Given information:
The two continuous random variables X and Y has joint density function defined as:
$f (x, y) = {\begin{cases} x + y o \leq x \leq 1, 0 \leq y \leq 1 \\ 0 o t h e r w i s e \end{cases}$
Then,
$P (X > x, Y > y) = \int_{y}^{1} \int_{x}^{1} f (x, y) d x d y$
Step 2
$P (X > 0.8, Y > 0.8) = \int_{y = 0.8}^{1} \int_{x = 0.8}^{1} (x + y) d x d y$
$= \int_{y = 0.8}^{1} {\int_{x = 0.8}^{1} (x + y) d x} d y$
$= \int_{y = 0.8}^{1} {[\frac{x^{2}}{2} + x y]}_{0.8}^{1} d y$
$= \int_{y = 0.8}^{1} {(\frac{1}{2} + y) - (\frac{{0.8}^{2}}{2} + 0.8 y)} d y$
$= \int_{y = 0.8}^{1} {\frac{1}{2} + y - 0.32 - 0.8} d y$
$= \int_{y = 0.8}^{1} (0.18 + 0.2 y) d y$
$= {[0.18 y + \frac{0.2 y^{2}}{2}]}_{0.8}^{1}$
$= [(0.18 + 0.1) - (0.18 \times 0.8 + 0.1 \times {0.8}^{2})]$
$= 0.072$
The required probability is 0.072.

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