390 Chapter 9: Regression
(b)Proof that Var
√
Y≈.25 whenYis Poisson with meanλ. Consider the Taylor series
expansion ofg(y) =√yabout the valueλ. By ignoring all terms beyond the second
derivative term, we obtain that
g(y)≈g(λ)+g′(λ)(y−λ)+
g′′(λ)(y−λ)^2
2
(9.8.2)
Since
g′(λ)=^12 λ−1/2, g′′(λ)=−^14 λ−3/2
we obtain, on evaluating Equation 9.8.2 aty=Y, that
√
Y≈
√
λ+^12 λ−1/2(Y−λ)−^18 λ−3/2(Y−λ)^2
Taking expectations, and using the results that
E[Y−λ]=0, E[(Y−λ)^2 ]=Var(Y)=λ
yields that
E[
√
Y]≈
√
λ−
1
8
√
λ
Hence
(E[
√
Y])^2 ≈λ+
1
64 λ
−
1
4
≈λ−
1
4
and so
Var(
√
Y)=E[Y]−(E[
√
Y])^2
≈λ−
(
λ−
1
4
)
=
1
4