Let (๐‘‹,๐‘Œ) be a pair of random variables with joint density โ„Ž(๐‘ฅ,๐‘ฆ)=๐‘ฅ+๐‘ฆ over the space [0,1]2.

1 answer

To find the marginal density functions for ๐‘‹ and ๐‘Œ, we need to integrate the joint density โ„Ž(๐‘ฅ,๐‘ฆ) over the appropriate ranges.

The marginal density function of ๐‘‹ is found by integrating โ„Ž(๐‘ฅ,๐‘ฆ) with respect to ๐‘ฆ over the entire range of ๐‘ฆ, which in this case is [0,1]:

๐‘“๐‘‹(๐‘ฅ) = โˆซ[0,1] (๐‘ฅ+๐‘ฆ) d๐‘ฆ

Simplifying the integral, we have:

๐‘“๐‘‹(๐‘ฅ) = ๐‘ฅโˆซ[0,1] d๐‘ฆ + โˆซ[0,1] ๐‘ฆ d๐‘ฆ
= ๐‘ฅ[y]โ‚€ยน + [yยฒ/2]โ‚€ยน
= ๐‘ฅ(1-0) + (1/2-0)
= ๐‘ฅ + 1/2

Therefore, the marginal density function of ๐‘‹ is ๐‘“๐‘‹(๐‘ฅ) = ๐‘ฅ + 1/2.

Similarly, the marginal density function of ๐‘Œ is found by integrating โ„Ž(๐‘ฅ,๐‘ฆ) with respect to ๐‘ฅ over the entire range of ๐‘ฅ, which is also [0, 1]:

๐‘“๐‘Œ(๐‘ฆ) = โˆซ[0,1] (๐‘ฅ+๐‘ฆ) d๐‘ฅ

Again, simplifying the integral, we have:

๐‘“๐‘Œ(๐‘ฆ) = โˆซ[0,1] ๐‘ฅ d๐‘ฅ + ๐‘ฆโˆซ[0,1] d๐‘ฅ
= [๐‘ฅยฒ/2]โ‚€ยน + ๐‘ฆ[x]โ‚€ยน
= (1/2-0) + ๐‘ฆ(1-0)
= 1/2 + ๐‘ฆ

Therefore, the marginal density function of ๐‘Œ is ๐‘“๐‘Œ(๐‘ฆ) = 1/2 + ๐‘ฆ.