Mean Squared Error of Variance Estimators
by Jeffrey S. Rosenthal, University of Toronto
Let x1 , x2 , . . . , xn be i.i.d., each with finite mean m and variance v =
E[(xi − m)2 ] and fourth central moment w = E[(xi − m)4 ].
P
Let B = a1 ni=1 (xi − x)2 be an estimator of v, for some fixed a > 0.
We are interested (because of the short paper [3]) in the Mean Squared
Error (MSE) when B is used as an estimator for the variance v. A formula
is claimed in [4], and related calculations appear in e.g. [1, 2]. Here for
completeness we derive a formula for M SE(B), making use of some related
moment calculations by Cramér [1].
I thank Mike Evans for helpful suggestions.
Proposition. The MSE of B as an estimator for v is given by
M SE(B) := E[(B − v)2 ] =
where γ =
w
v2
i
h 2(n − 1)
i
n − 1h
2
2
(n
−
1)γ
+
n
+
n
v
−
−
1
v2 ,
na2
a
− 3 the excess kurtosis (so w = v 2 (γ + 3)).
Proof. Let µν = E[(xi − m)ν ], and let mν = n1 ni=1 (xi − x)ν . Then our
P
estimator is B = a1 ni=1 (xi − x)2 = (n/a)m2 . We want the MSE of B, i.e.
P
M SE(B) := E[(B − v)2 ] = E[B 2 ] − 2 v E[B] + v 2
Now, Cramér equation (27.4.1) on page 347 says that E[m2 ] = n−1
µ2
n
where µ2 = E[(xi − m)2 ] = v. Hence, E[B] = (n/a)E[m2 ] = n−1
v.
Also,
a
Cramér equation (27.4.2) on page 348 says that
E[(m2 )2 ] = µ22 +
µ4 − 3µ22 2µ4 − 5µ22 µ4 − 3µ22
−
+
,
n
n2
n3
where µ2 = v as above, and where µ4 = E[(xi − m)4 ] = w. Hence,
"
E[B 2 ] = (n/a)2 E[(m2 )2 ] = (n/a)2
µ4 − 3µ22 2µ4 − 5µ22 µ4 − 3µ22
µ22 +
−
+
.
n
n2
n3
#
Putting this all together,
M SE(B) = E[B 2 ] − 2 v E[B] + v 2
1
µ4 − 3µ22 2µ4 − 5µ22 µ4 − 3µ22 i
n−1
−
v + v2 .
+
− 2v
2
3
n
n
n
a
Substituting in µ2 = v and µ4 = w = v 2 (γ + 3), this becomes
h
= (n/a)2 µ22 +
h
M SE(B) = (n/a)2 v 2 +
v 2 (γ + 3) − 3v 2 2v 2 (γ + 3) − 5v 2 v 2 (γ + 3) − 3v 2 i
n−1
−
v+v 2
+
−2v
2
3
n
n
n
a
i
γ 2γ + 1 v 2 γ i h 2(n − 1)
−
−
1
v2
+
−
n
n2
n3
a
h
i
1
1
2
1 i h 2(n − 1)
= (nv/a)2 (1 + 2 ) + γ( − 2 + 3 ) −
− 1 v2
n
n n
n
a
i
h n2 + 1
n2 − 2n + 1 i h 2(n − 1)
= (nv/a)2
+
γ
−
1
v2
−
2
3
n
n
a
2i
h
i
h
(n − 1)
2(n − 1)
= (v/a)2 (n + 1)(n − 1) + γ
− 1 v2
−
n
a
h
i
h
i
2(n − 1)
n−1
2
n(n
+
1)
+
γ(n
−
1)
v
−
−
1
v2
=
na2
a
i
h 2(n − 1)
i
n − 1h
2
2
=
(n
−
1)γ
+
n
+
n
v
−
−
1
v2 .
Q.E.D.
na2
a
h
= (nv/a)2 1 +
References
[1] H. Cramér (1946), Mathematical Methods of Statistics. Princeton University Press.
[2] A. Mood, F. Graybill, and D. Boes (1974), Introduction to the Theory
of Statistics (3rd ed.), p. 229. McGraw-Hill, New York City.
[3] J.S. Rosenthal (2015), The kids are alright: divide by n when estimating
variance. IMS Bulletin, to appear.
[4] Wikipedia, Mean squared error: Variance. Retrieved August 26, 2015.
Available at:
en.wikipedia.org/wiki/Mean squared error#Variance
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