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Bivariate Noremal Distribution
Let the pair (𝑋, π‘Œ) of random variables follow a bivariate normal distribution. Then the probability
density function of (𝑋, π‘Œ) is
𝑓𝑋,π‘Œ (π‘₯, 𝑦) =
1
2πœ‹πœŽ1 𝜎2 √1 βˆ’ 𝜌2
𝑒
βˆ’
1
π‘₯βˆ’πœ‡1 2
π‘₯βˆ’πœ‡1 π‘¦βˆ’πœ‡2
π‘¦βˆ’πœ‡2 2
[(
) βˆ’2𝜎(
)(
)+(
) ]
𝜎1
𝜎1
𝜎2
𝜎2
2(1βˆ’πœŒ2 )
βˆ’βˆž < π‘₯ < ∞, βˆ’βˆž < 𝑦 < ∞, βˆ’βˆž < πœ‡1 < ∞, βˆ’βˆž < πœ‡2 < ∞, βˆ’1 < 𝜌 < 1.
The marginal density function of 𝑋 is
∞
𝑓𝑋 (π‘₯) = ∫ 𝑓𝑋,π‘Œ (π‘₯, 𝑦) 𝑑𝑦
βˆ’βˆž
∞
= ∫
βˆ’βˆž
∞
= ∫(
βˆ’βˆž
=(
1
2πœ‹πœŽ1 𝜎2 √1 βˆ’ 𝜌2
1
√2πœ‹πœŽ1
1
√2πœ‹πœŽ1
𝑒
βˆ’
𝑒
βˆ’
1
(π‘₯βˆ’πœ‡1 )2
2𝜎1 2
)(
1
(π‘₯βˆ’πœ‡1 )2
βˆ’
𝑒 2𝜎1 2
)
=
1
√2πœ‹πœŽ1
𝑒
1
π‘₯βˆ’πœ‡1 2
π‘₯βˆ’πœ‡1 π‘¦βˆ’πœ‡2
π‘¦βˆ’πœ‡2 2
[(
) βˆ’2𝜎(
)(
)+(
) ]
𝜎1
𝜎1
𝜎2
𝜎2
2(1βˆ’πœŒ2 )
𝑑𝑦
2
1
√2πœ‹(1 βˆ’ 𝜌2 𝜎2
∞
∫
βˆ’
1
√2πœ‹(1 βˆ’ 𝜌2 𝜎2
βˆ’βˆž
βˆ’
1
(π‘₯βˆ’πœ‡1 )2
2𝜎1 2
𝑒
βˆ’
×1 =
𝑒
1
𝜎
((π‘¦βˆ’πœ‡2 )βˆ’πœŒπœŽ2 (π‘₯βˆ’πœ‡1 ))
2(1βˆ’πœŒ2 )𝜎2 2
1
1
𝜎
(π‘¦βˆ’(πœ‡2 +𝜌 2 (π‘₯βˆ’πœ‡1 )))
𝜎1
2(1βˆ’πœŒ2 )𝜎2 2
1
√2πœ‹πœŽ1
𝑒
βˆ’
) 𝑑𝑦
2
𝑑𝑦
1
(π‘₯βˆ’πœ‡1 )2
2𝜎1 2
(The integrand under the last integral is the probability density function of a normal distribution with
𝜎
mean πœ‡2 + 𝜌 𝜎2 (π‘₯ βˆ’ πœ‡1 ) and variance (1 βˆ’ 𝜌2 )𝜎2 2. So the integral is equal to 1.)
1
So the marginal density function of 𝑋 is
1
𝑓𝑋 (π‘₯) =
(π‘₯βˆ’πœ‡1 )2
βˆ’
1
2𝜎1 2
𝑒
, βˆ’βˆž
√2πœ‹πœŽ1
< π‘₯ < ∞, βˆ’βˆž < πœ‡1 < ∞.
Similarly, the marginal density function of Y is
1
π‘“π‘Œ (𝑦) =
(π‘₯βˆ’πœ‡2 )2
βˆ’
1
𝑒 2𝜎22
, βˆ’βˆž
√2πœ‹πœŽ2
< 𝑦 < ∞, βˆ’βˆž < πœ‡2 < ∞.
So X and Y are normally distributed.
𝐸(𝑋) = πœ‡1 , π‘£π‘Žπ‘Ÿ(𝑋) = 𝜎1 2
𝐸(π‘Œ) = πœ‡2 , π‘£π‘Žπ‘Ÿ(π‘Œ) = 𝜎2 2
Consider the transformation
π‘ˆ =π‘‹βˆ’πœŒ
𝜎1
π‘Œ,
𝜎2
𝑋 =π‘ˆ+𝜌
𝑉=π‘Œ
𝜎1
𝑉, π‘Œ = 𝑉
𝜎2
The jacobian of transformation is
πœ•π‘₯
𝐽 = |πœ•π‘’
πœ•π‘¦
πœ•π‘’
πœ•π‘₯
πœ•π‘£ | = |1
πœ•π‘¦
0
πœ•π‘£
𝜌
𝜎1
𝜎2 | = 1
1
The joint density of π‘ˆ π‘Žπ‘›π‘‘ 𝑉 is
π‘“π‘ˆ,𝑉 (𝑒, 𝑣) = 𝑓𝑋,π‘Œ (π‘₯, 𝑦) × |𝐽| =
=
=
1
√2πœ‹πœŽ2
1
√2πœ‹πœŽ2
=
𝑒
βˆ’
𝑒
βˆ’
√2πœ‹πœŽ2
2πœ‹πœŽ1 𝜎2 √1 βˆ’ 𝜌2
1
(π‘¦βˆ’πœ‡2 )2
2𝜎2 2
1
(π‘£βˆ’πœ‡2 )2
2𝜎2 2
1
1
𝑒
βˆ’
𝑒
βˆ’
1
π‘₯βˆ’πœ‡1 2
π‘₯βˆ’πœ‡1 π‘¦βˆ’πœ‡2
π‘¦βˆ’πœ‡2 2
[(
) βˆ’2𝜎(
)(
)+(
) ]
𝜎1
𝜎1
πœ‡2
πœ‡2
2(1βˆ’πœŒ2 )
2
×
1
√2πœ‹(1 βˆ’ 𝜌2 𝜎1
𝑒
βˆ’
1
𝜎
((π‘₯βˆ’πœ‡1 )βˆ’πœŒπœŽ1 (π‘¦βˆ’πœ‡2 ))
2(1βˆ’πœŒ2 )𝜎1 2
2
2
×
1
√2πœ‹(1 βˆ’ 𝜌2 𝜎1
1
(π‘£βˆ’πœ‡2 )2
2𝜎2 2
×
𝑒
βˆ’
1
𝜎
𝜎
((𝑒+𝜌𝜎1 π‘£βˆ’πœ‡1 )βˆ’πœŒπœŽ1 (π‘£βˆ’πœ‡2 ))
2(1βˆ’πœŒ2 )𝜎1 2
2
2
1
√2πœ‹(1 βˆ’ 𝜌2 𝜎1
𝑒
βˆ’
2
1
𝜎
(π‘’βˆ’(πœ‡1 βˆ’πœŒ 1 πœ‡2 ))
𝜎2
2(1βˆ’πœŒ2 )𝜎1 2
This shows that
𝜎
(1) π‘ˆ = 𝑋 βˆ’ 𝜌 𝜎1 π‘Œ and 𝑉 = π‘Œ are independently distributed
2
(2) Y=V follows normal distribution with mean πœ‡2 and variance 𝜎2 2
𝜎
(3) The conditional distribution of π‘ˆ = 𝑋 βˆ’ 𝜌 𝜎1 π‘Œ given V= π‘Œ = 𝑦 is normal with mean πœ‡1 βˆ’
𝜎
𝜌 𝜎1 πœ‡2
2
2
and variance (1
βˆ’ 𝜌2 )𝜎1 2.
From (3) it follows that the conditional distribution of 𝑋 given π‘Œ = 𝑦 is normal with mean πœ‡1 +
𝜎
𝜌 1 (𝑦 βˆ’ πœ‡2 ) and variance (1 βˆ’ 𝜌2 )𝜎1 2 .
𝜎2
Distribution of 𝑿 + 𝒀
𝑋+π‘Œ =π‘ˆ+𝜌
𝜎1
𝜎1
𝑉 + 𝑉 = π‘ˆ + (1 + 𝜌 ) 𝑉
𝜎2
𝜎2
Since 𝑋 + π‘Œ is a linear combination of the independent normal random variables π‘ˆ and 𝑉, the
distribution of X+Y is normal with mean 𝐸(𝑋 + π‘Œ) = 𝐸(𝑋) + 𝐸(π‘Œ) = πœ‡1 + πœ‡2 and variance
𝜎
𝜎
2
𝜎
π‘‰π‘Žπ‘Ÿ(𝑋 + π‘Œ) = π‘‰π‘Žπ‘Ÿ (π‘ˆ + (1 + 𝜌 𝜎1 ) 𝑉) = π‘‰π‘Žπ‘Ÿ(π‘ˆ) + (1 + 𝜌 𝜎1 ) π‘‰π‘Žπ‘Ÿ(𝑉) + 2 (1 + 𝜌 𝜎1 ) πΆπ‘œπ‘£(π‘ˆ, 𝑉)
2
𝜎
2
= (1 βˆ’ 𝜌2 )𝜎1 2 + (1 + 𝜌 𝜎1 ) 𝜎2 2 = 𝜎1 2 + 𝜎2 2 + 2𝜌𝜎1 𝜎2.
2
2
2
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