Introduction to Probability and Statistics for Engineers and Scientists

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