Given that F X , Y ( x , y ) = k ( 4 x 2 y 2 + 5 x y 4 ) , 0 < x < 1 , 0 < y < 1 , find the corresponding pdf and use it to calculate P ( 0 < X < 1 2 , 1 2 < Y < 1 ) .
Given that F X , Y ( x , y ) = k ( 4 x 2 y 2 + 5 x y 4 ) , 0 < x < 1 , 0 < y < 1 , find the corresponding pdf and use it to calculate P ( 0 < X < 1 2 , 1 2 < Y < 1 ) .
Solution Summary: The author explains that the joint pdf of X and Y, f_X,Y(x,y), is a second partial derivative of this joint Cdf.
Given that
F
X
,
Y
(
x
,
y
)
=
k
(
4
x
2
y
2
+
5
x
y
4
)
,
0
<
x
<
1
,
0
<
y
<
1
, find the corresponding pdf and use it to calculate
P
(
0
<
X
<
1
2
,
1
2
<
Y
<
1
)
.
Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.
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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY