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The joint pdf of X and Y is given by f(x,y)=k(x2+2xy​),x∈(0,1),y∈(0,2). a. Find the constant k so that f is a legitimate joint pdf. b Find P(Y

User Mirlande
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The function f(x, y) is defined as k * (x^2 + 2*x*y), where k is a normalization constant to ensure that the function f(x, y) is a valid probability density function (pdf). The limits of x are (0, 1) and for y are (0, 2).

a. To make f(x, y) be a valid pdf, the double integral of f(x, y) over its entire range (0 ≤ x ≤ 1, 0 ≤ y ≤ 2) must equal 1.

Therefore, we have the equation:

∬f(x, y) dx dy = 1

Substitute f(x, y) = k * (x^2 + 2*x*y):

∬ [k * (x^2 + 2*x*y)] dx dy = 1

Take the integral part by part:

∫ from 0 to 2 [∫ from 0 to 1 [k * (x^2 + 2*x*y)] dx ] dy = 1

After performing the double integration, equate it to 1. Solve the resulting equation for k. This will give you the constant k that ensures f(x, y) is a valid pdf.

b. To find P(Y < 1 | X = 0.5), we first need to condition on the value X = 0.5. This results in the conditional pdf f(y|0.5) where x = 0.5 in f(x, y).

Now, to find P(Y < 1 | X = 0.5), we integrate this conditional pdf from 0 to 1. This single integral of the conditional pdf over the range (0, 1) will give you the probability P(Y < 1 | X = 0.5).

User Regmagik
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