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Suppose that the joint PDF of (๐‘‹, ๐‘Œ) is๐‘“(๐‘ฅ, ๐‘ฆ) = ๐‘โˆš1 โˆ’ ๐‘ฅ2 โˆ’ ๐‘ฆ2, ๐‘ฅ2 + ๐‘ฆ2 โ‰ค 1.Find the marginal PDF ๐‘“๐‘‹(๐‘ฅ) and the constant ๐‘ . (iint: consider transformations like๐‘ฆ = ๐‘Ž sin(๐œƒ) when calculating the integral) (10 points)

Question

Suppose that the joint PDF of (๐‘‹, ๐‘Œ) is๐‘“(๐‘ฅ, ๐‘ฆ) = ๐‘โˆš1 โˆ’ ๐‘ฅ2 โˆ’ ๐‘ฆ2, ๐‘ฅ2 + ๐‘ฆ2 โ‰ค 1.Find the marginal PDF ๐‘“๐‘‹(๐‘ฅ) and the constant ๐‘ . (iint: consider transformations like๐‘ฆ = ๐‘Ž sin(๐œƒ) when calculating the integral) (10 points)

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Solution

To find the marginal PDF ๐‘“๐‘‹(๐‘ฅ) and the constant ๐‘, we need to integrate the joint PDF over the entire range of y.

First, let's find the constant ๐‘. The joint PDF must integrate to 1 over the region ๐‘ฅยฒ + ๐‘ฆยฒ โ‰ค 1. This is a circle of radius 1, so we can switch to polar coordinates for easier calculation. Let ๐‘ฅ = ๐‘Ÿ cos(๐œƒ) and ๐‘ฆ = ๐‘Ÿ sin(๐œƒ). Then ๐‘Ÿยฒ = ๐‘ฅยฒ + ๐‘ฆยฒ and ๐‘‘๐‘ฅ๐‘‘๐‘ฆ = ๐‘Ÿ๐‘‘๐‘Ÿ๐‘‘๐œƒ. The limits of integration for ๐‘Ÿ are 0 to 1 and for ๐œƒ are 0 to 2๐œ‹.

So, we have:

โˆซโˆซ ๐‘โˆš1 - ๐‘Ÿยฒ * ๐‘Ÿ๐‘‘๐‘Ÿ๐‘‘๐œƒ from ๐‘Ÿ=0 to 1 and ๐œƒ=0 to 2๐œ‹ = 1

Solving this integral gives ๐‘ = 1/2๐œ‹.

Now, let's find the marginal PDF ๐‘“๐‘‹(๐‘ฅ). This is the integral of the joint PDF over all ๐‘ฆ, or in polar coordinates, over all ๐œƒ from 0 to 2๐œ‹.

So, ๐‘“๐‘‹(๐‘ฅ) = โˆซ ๐‘โˆš1 - ๐‘ฅยฒ - ๐‘ฆยฒ ๐‘‘๐‘ฆ from ๐‘ฆ=-โˆš(1-๐‘ฅยฒ) to โˆš(1-๐‘ฅยฒ)

Switching to polar coordinates, this becomes:

๐‘“๐‘‹(๐‘ฅ) = โˆซ ๐‘โˆš1 - ๐‘Ÿยฒ * ๐‘Ÿ๐‘‘๐œƒ from ๐œƒ=0 to 2๐œ‹

Solving this integral gives ๐‘“๐‘‹(๐‘ฅ) = โˆš(1 - ๐‘ฅยฒ) for -1 โ‰ค ๐‘ฅ โ‰ค 1.

This problem has been solved

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