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We determined that

f(y₁, y₂) = {6(1 − y₂), 0 ≤ y₁ ≤ y₂ ≤ 1,
{0, elsewhere
is a valid joint probability density function.

(a) Find E(Y₁|Y₂ = y₂).

User Neilski
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1 Answer

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Final answer:

To find E(Y₁|Y₂ = y₂), we can use the conditional probability density function f(y₁|y₂) and calculate the expected value of Y₁.

Step-by-step explanation:

To find E(Y₁|Y₂ = y₂), we need to calculate the conditional expected value of Y₁ given Y₂ = y₂.

From the joint probability density function, we have f(y₁, y₂) = 6(1 - y₂) for 0 ≤ y₁ ≤ y₂ ≤ 1.

The conditional probability density function can be expressed as f(y₁|y₂) = f(y₁, y₂)/f(y₂).

To find f(y₂), we integrate f(y₁, y₂) over the range of y₁ from 0 to y₂.

Integrating f(y₁, y₂) = 6(1 - y₂) from 0 to y₂ with respect to y₁, we get F(y₂) = 6y₂ - 3y₂².

Now, we can find f(y₁|y₂) by differentiating F(y₂) with respect to y₂.

f(y₁|y₂) = d/dy₂(F(y₂)) = 6 - 6y₂.

Finally, we can calculate E(Y₁|Y₂ = y₂) by finding the expected value of Y₁ using the conditional probability density function f(y₁|y₂).

User Steve Lovell
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