Proposition. Suppose X1 ∼ N (µ1 , σ12 ) and X2 ∼ N (µ2 , σ22 ) are independent, then
X1 + X2 ∼ N (µ1 + µ2 , σ12 + σ22 ).
Proof. Let us write Z = X1 + X2 . Then, as shown previously,
Z ∞
fX1 (z − y)fX2 (y) dy
fZ (z) =
−∞
Z ∞
1
(z − y − µ1 )2 (y − µ2 )2
=
dy
exp −
−
2πσ1 σ2 −∞
2σ12
2σ22
To proceed, we must complete the square in the exponent. In general we have
b 2
b2
ay 2 + by + c = a y + 2a
+ c − 4a
.
In our particular case,
(z − y − µ1 )2 (y − µ2 )2
+
σ12
σ2
2
1
1
z − µ1 µ2
(z − µ1 )2 (µ2 )2
2
= 2 + 2 y −2
+ 2 y+
+ 2
σ1 σ1
σ12
σ2
σ12
σ2
2 2
2
2 2
σ1 + σ2
σ1 σ2
z − µ1 µ2
(z − µ1 )2 (µ2 )2
2
= 2 2 (y − y0 ) − 2
+ 2 +
+ 2
σ1 σ2
σ1 + σ22
σ12
σ2
σ12
σ2
where
σ12 σ22
z − µ1 µ2
y0 = 2
+ 2
σ1 + σ22
σ12
σ2
As we will see, there is no advantage to simplifying the formula for y0 ; however we
do need to simplify the last two terms. After some effort, we find
σ12 + σ22
(z − µ1 − µ2 )2
(z − y − µ1 )2 (y − µ2 )2
2
+
=
(y
−
y
)
+
.
0
σ12
σ22
σ12 σ22
σ12 + σ22
Consequently,
2
Z ∞
σ1 + σ22
(z − µ1 − µ2 )2
1
2
exp −
(y − y0 ) −
fZ (z) =
dy
2πσ1 σ2 −∞
2σ12 σ22
2(σ12 + σ22 )
Now, recalling the general result
Z ∞
√
(y − y0 )2
exp −
dy
=
2πσ 2 ,
2
2σ
−∞
which was shown in class, we deduce that
s
σ12 σ22
1
(z − µ1 − µ2 )2
fZ (z) =
exp −
2π
2πσ1 σ2
2(σ12 + σ22 )
σ12 + σ22
1
(z − µ1 − µ2 )2
=p
exp −
.
2(σ12 + σ22 )
2π(σ12 + σ22 )
This then shows that Z ∼ N (µ1 + µ2 , σ12 + σ22 ).
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