Statistics Graduate Level Quiz: Influence Functions

10 questions · exam conditions

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Influence FunctionsQuestion 1 of 10

Let $T(F)$ be the median of a distribution $F$ having a unique median $m$ and a density continuous and positive at $m$ . Suppose $f(m)=0.20$ . A small contamination mass is placed at a point $z>m$ .

For $F_{\varepsilon}=(1-\varepsilon)F+\varepsilon\Delta_z$ , what is the influence function $\operatorname{IF}(z;T,F)$ ?

-2.5

, because contamination above the median shifts the distribution function downward at the original median.

2.5

, because the median must move upward to restore cumulative probability one-half.

5.0

, because the contamination effect is scaled by the reciprocal of the density at the median.

0.10

, because half of the contamination mass contributes to the first-order median shift.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Influence Functions

Practice Influence Functions in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Influence Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let $T(F)$ be the median of a distribution $F$ having a unique median $m$ and a density continuous and positive at $m$ . Suppose $f(m)=0.20$ . A small contamination mass is placed at a point $z>m$ .

For $F_{\varepsilon}=(1-\varepsilon)F+\varepsilon\Delta_z$ , what is the influence function $\operatorname{IF}(z;T,F)$ ?

$-2.5$ , because contamination above the median shifts the distribution function downward at the original median.
$2.5$ , because the median must move upward to restore cumulative probability one-half. (correct answer)
$5.0$ , because the contamination effect is scaled by the reciprocal of the density at the median.
$0.10$ , because half of the contamination mass contributes to the first-order median shift.

Explanation: Whenever you encounter influence function questions, your anchor should be the formal definition:

\operatorname{IF}(z; T, F) = \lim_{\varepsilon \to 0} \frac{T(F_\varepsilon) - T(F)}{\varepsilon}

, which measures the first-order sensitivity of a functional to infinitesimal contamination at point

z

. For the sample median, the derivation proceeds by solving for where the contaminated distribution

F_\varepsilon = (1-\varepsilon)F + \varepsilon\Delta_z

achieves cumulative probability one-half. Setting

F_\varepsilon(m_\varepsilon) = 1/2

and differentiating implicitly with respect to

\varepsilon

\varepsilon = 0

yields the closed-form result:

\operatorname{IF}(z; T, F) = \frac{\operatorname{sgn}(z - m)}{2f(m)}

Since

z > m

\operatorname{sgn}(z - m) = +1

, so

\operatorname{IF}(z; T, F) = \frac{1}{2 \times 0.20} = \frac{1}{0.40} = 2.5

. The median must shift upward to compensate for the extra mass above it — confirming B is correct. Choice A gets the sign wrong. Contamination above the median pushes the median upward, not downward; the distribution function at the original median increases, forcing the quantile to move right, not left. Choice C arrives at

5.0

by using

1/f(m)

alone, forgetting the factor of

2

in the denominator — a classic algebra slip. Choice D produces

0.10

by multiplying

\varepsilon/2

thinking proportionally, which confuses a finite-sample heuristic with the correct asymptotic derivative formula. Your study tip: always memorize the median's influence function as

\frac{\operatorname{sgn}(z-m)}{2f(m)}

. The factor of 2 reflects the two-sided nature of the median condition, and

f(m)

in the denominator means the influence is larger when the density is flat — intuition worth retaining.

Question 2

Define $T(F)=E_F(X^2)/E_F(X)$ , where $E_F(X)=2$ and $E_F(X^2)=5$ . Consider contamination at $z=4$ .

Using the first-order contamination derivative, what is $\operatorname{IF}(4;T,F)$ ?

$1.5$ , because only the contamination effect on the numerator remains after centering.
$8.0$ , because the contamination contributes its squared value relative to the first moment.
$5.5$ , because the numerator derivative is divided by the original first moment.
$3.0$ , because both the numerator and denominator change under contamination. (correct answer)

Explanation: When you see a functional of the form

T(F) = E_F(X^2)/E_F(X)

, recognize it as a ratio functional. The influence function requires applying the quotient rule to the contaminated functional

T(F_\epsilon) = E_{F_\epsilon}(X^2)/E_{F_\epsilon}(X)

, where

F_\epsilon = (1-\epsilon)F + \epsilon\delta_z

. Under contamination by a point mass at

z

, the numerator becomes

(1-\epsilon)E_F(X^2) + \epsilon z^2

and the denominator becomes

(1-\epsilon)E_F(X) + \epsilon z

. Differentiating

T(F_\epsilon)

with respect to

\epsilon

\epsilon=0

via the quotient rule gives:

\operatorname{IF}(z;T,F) = \frac{(z^2 - E_F(X^2)) \cdot E_F(X) - E_F(X^2)\cdot(z - E_F(X))}{[E_F(X)]^2}

Plugging in

z=4

E_F(X)=2

E_F(X^2)=5

= \frac{(16-5)\cdot 2 - 5\cdot(4-2)}{4} = \frac{22 - 10}{4} = \frac{12}{4} = 3.0

This confirms D is correct. Choice A incorrectly ignores the denominator's contribution entirely. Choice B simply reports

z^2/E_F(X) = 8

, confusing the raw contaminated value with the influence function derivative. Choice C computes only

(z^2 - E_F(X^2))/E_F(X) = 11/2 = 5.5

, applying only the numerator's centered derivative without accounting for the denominator's shift. As a study rule: whenever

T(F)

is a ratio, always apply the full quotient rule for the IF — forgetting the denominator's derivative is the single most common error on these problems.

Question 3

A location functional $T(F)$ is defined by $E_F\{\psi(X-T(F))\}=0$ , where $\psi(u)=\max(-1.5,\min(u,1.5))$ . At the model distribution, $T(F)=0$ , $X$ is standard normal, and $P_F(|X|\le 1.5)=0.8664$ .

Approximately what is the influence function at a contamination point $z=4$ ?

$0.578$ , obtained by dividing the expected derivative by the clipped contamination score.
$1.500$ , obtained by using the clipped score without accounting for sensitivity.
$1.731$ , obtained by scaling the clipped score by the inverse expected derivative. (correct answer)
$4.617$ , obtained by scaling the unclipped residual by the inverse expected derivative.

Explanation: When you encounter influence function (IF) questions, the key formula to internalize is:

\text{IF}(z; T, F) = \frac{\psi(z - T(F))}{E_F[\psi'(X - T(F))]}

This comes from differentiating the estimating equation

E_F[\psi(X - T(F))] = 0

with respect to contamination at point

z

. The numerator captures how the score function evaluates the outlier; the denominator measures the estimator's average sensitivity. At

T(F) = 0

and

z = 4

, the numerator is

\psi(4) = \min(4, 1.5) = 1.5

— the score is clipped. For the denominator, since

\psi

is the Huber-type clip function,

\psi'(u) = 1

when

|u| \le 1.5

and

0

otherwise. So

E_F[\psi'(X)] = P_F(|X| \le 1.5) = 0.8664

. The influence function is therefore:

\text{IF}(4) = \frac{1.5}{0.8664} \approx 1.731

This confirms C is correct. A is wrong because it inverts the formula — dividing the expected derivative by the score, rather than the score by the expected derivative. B simply reads off the clipped score (1.5) and ignores the denominator entirely, failing to account for how steeply the estimator responds to perturbations. D uses the raw, unclipped value

z = 4

in the numerator, as if

\psi

were the identity function — a fundamental misreading of what the redescending/bounded

\psi

does to outliers. Study tip: For M-estimator influence functions, always write down the estimating equation first, then differentiate. The denominator

E[\psi']

is never 1 unless

\psi

is the identity — don't skip it.

Question 4

Two location estimators are under consideration. The first has a bounded influence function at the model distribution, whereas the second has an unbounded influence function. No additional information about their finite-sample behavior is available.

Which conclusion is justified solely from this information?

The first estimator has a positive finite-sample breakdown point, whereas the second estimator has breakdown point zero.
The first estimator has bounded first-order sensitivity to infinitesimal point-mass contamination at the model. (correct answer)
The first estimator has smaller asymptotic variance than the second estimator under every distribution in the model.
The first estimator remains uniformly accurate under contamination by any fixed positive fraction of arbitrary observations.

Explanation: When studying robustness theory, the key is distinguishing precisely what each concept measures. The influence function (IF) quantifies the asymptotic effect of an infinitesimal point-mass contamination on an estimator's value. Formally, if

T

is a functional and

F

is the model distribution, the IF at point

x

is:

\text{IF}(x; T, F) = \lim_{\epsilon \to 0} \frac{T((1-\epsilon)F + \epsilon\delta_x) - T(F)}{\epsilon}

A bounded influence function means this derivative is uniformly bounded over all contamination points

x

, which is precisely a statement about first-order sensitivity to infinitesimal contamination. This is exactly what answer B states — it's a direct, definitional consequence of having a bounded IF, making B correct. Answer A is tempting but unjustified. Bounded IF and positive breakdown point are related robustness ideas, but neither implies the other. An estimator can have a bounded IF yet still have breakdown point zero (e.g., certain M-estimators with bounded scores but non-robust scale choices). The IF is a local, infinitesimal concept; breakdown point is a global, finite-sample concept. You cannot infer one from the other without additional information. Answer C is false — boundedness of the IF says nothing about asymptotic variance. In fact, high-breakdown estimators often trade efficiency for robustness, meaning the first estimator could have larger variance. Answer D overreaches significantly. A bounded IF only guarantees infinitesimal-contamination stability; it provides no uniform guarantee over a fixed positive contamination fraction

\epsilon > 0

. Study tip: Always match robustness concepts to their precise scope — IF is infinitesimal and asymptotic, breakdown point is finite and global. Exam distractors frequently blur this boundary.

Question 5

Let $T(F)=\{E_F(X)\}^2$ , and suppose the true distribution has $E_F(X)=0$ and $0<\operatorname{Var}_F(X)=\sigma^2<\infty$ . The plug-in estimator is $T(F_n)=\bar X_n^2$ .

Which statement correctly describes the influence function and the first nondegenerate asymptotic behavior of this estimator?

The influence function is zero, and $n\bar X_n^2$ converges in distribution to $\sigma^2\chi_1^2$ . (correct answer)
The influence function is $2X$ , and $\sqrt n\,\bar X_n^2$ converges to a centered normal distribution.
The influence function is $X^2-\sigma^2$ , and $\sqrt n\,\bar X_n^2$ has asymptotic variance $2\sigma^4$ .
The influence function is zero, and $n\bar X_n^2$ converges in probability to the constant $\sigma^2$ .

Explanation: When a functional

T(F)

is smooth, its plug-in estimator inherits a

\sqrt{n}

-rate via the delta method, driven by a nonzero influence function. But when the influence function vanishes, the estimator is second-order, and you need to look one level deeper. The influence function of

T(F) = \{E_F(X)\}^2

is the Gateaux derivative:

\text{IF}(x; T, F) = \frac{d}{d\epsilon}\Big|_{\epsilon=0} T((1-\epsilon)F + \epsilon\delta_x) = 2E_F(X)\cdot(x - E_F(X)).

Since

E_F(X) = 0

by assumption, the influence function equals zero. This means the standard

\sqrt{n}

central limit theorem gives a degenerate (zero) limit for

\sqrt{n}\,\bar{X}_n^2

, so you must rescale by

n

instead. By the CLT,

\sqrt{n}\,\bar{X}_n \xrightarrow{d} N(0, \sigma^2)

, and therefore

n\bar{X}_n^2 = \left(\sqrt{n}\,\bar{X}_n\right)^2 \xrightarrow{d} \sigma^2 \chi_1^2,

confirming answer A. Answer B is wrong on both counts: the influence function is not

2X

(that would require

E_F(X)\neq 0

), and

\sqrt{n}\,\bar{X}_n^2 \to 0

in probability, not a normal. Answer C confuses

T(F)

with the variance functional

\text{Var}_F(X) = E[X^2] - (E[X])^2

; the influence function

X^2 - \sigma^2

belongs to a different functional entirely. Answer D gets the rescaling right but claims convergence in probability to a constant — that would require a degenerate limit, but

\sigma^2\chi_1^2

is a genuine random variable, not a constant. Study tip: Whenever

E_F(X) = 0

makes the first-order influence function vanish, immediately switch to a second-order analysis — rescale by

n

, apply the continuous mapping theorem to the squared CLT limit, and expect a chi-squared (not normal) distribution.

Question 6

Let $X$ be Bernoulli with success probability $p=0.25$ . The parameter of interest is the log-odds functional $\theta(F)=\log\{p/(1-p)\}$ , where $p=E_F(X)$ .

What is the influence function for $\theta$ at the contamination point $z=1$ ?

$0.75$ , because this is the influence of a success on the probability functional.
$3.00$ , because the probability influence is divided only by the success probability.
$4.00$ , because the probability influence is scaled by the derivative of the log-odds. (correct answer)
$5.33$ , because the reciprocal Bernoulli variance is the influence of one success.

Explanation: Whenever you encounter influence function questions, your instinct should be to apply the chain rule through the statistical functional. The influence function of a transformed functional

\theta(F) = g(\mu(F))

is:

\text{IF}(z; \theta, F) = g'(\mu) \cdot \text{IF}(z; \mu, F)

Here,

\mu(F) = p = E_F[X]

and

g(p) = \log\{p/(1-p)\}

. The influence function for the mean functional at point

z

is simply

\text{IF}(z; \mu, F) = z - p

. At

z = 1

and

p = 0.25

, this equals

1 - 0.25 = 0.75

. The derivative of the log-odds is

g'(p) = 1/[p(1-p)]

. At

p = 0.25

g'(0.25) = 1/(0.25 \times 0.75) = 1/0.1875 \approx 5.33

. Multiplying gives

5.33 \times 0.75 = 4.00

, confirming C is correct. Choice A stops too early —

0.75

is the influence function for the mean functional, not the log-odds. It ignores the chain rule transformation entirely. Choice B computes

0.75 / 0.25 = 3.00

, dividing only by

p

rather than by the full Bernoulli variance

p(1-p)

, which misapplies the derivative of the log-odds. Choice D reports

5.33

, which is

g'(p)

alone — the derivative of the log-odds evaluated at

p

— but forgets to multiply by the mean's influence function

z - p = 0.75

. As a study tip: always decompose functional influence functions using the chain rule — identify the inner functional's IF first, then scale by the outer function's derivative. This two-step process prevents every trap present in this question.

Question 7

A regular estimator $T_n$ has the asymptotic linear representation $T_n-T(F)=n^{-1}\sum_{i=1}^n \operatorname{IF}(X_i;T,F)+o_p(n^{-1/2})$ . At the distribution of interest, $\operatorname{IF}(X;T,F)=X^2-2$ , $E_F(X^2)=2$ , and $E_F(X^4)=10$ .

What is the asymptotic variance of $\sqrt{n}\{T_n-T(F)\}$ ?

$2$ , because the influence function is centered by the second moment.
$10$ , because the fourth moment determines the limiting variance directly.
$8$ , because the cross term contributes twice the squared second moment.
$6$ , because the limiting variance is the second moment of the influence function. (correct answer)

Explanation: Whenever you encounter an asymptotic linear representation, your first instinct should be: the asymptotic variance of

\sqrt{n}(T_n - T(F))

equals the variance of the influence function. By the CLT applied to the i.i.d. sum,

\sqrt{n}(T_n - T(F)) \xrightarrow{d} N(0, \text{Var}_F(\operatorname{IF}(X;T,F)))

. So the task reduces to computing

\text{Var}_F(X^2 - 2)

. Since

E_F(X^2) = 2

, the influence function

\operatorname{IF}(X;T,F) = X^2 - 2

is already mean-zero (a requirement for valid influence functions). Its variance is:

\text{Var}_F(X^2 - 2) = E_F[(X^2-2)^2] = E_F[X^4 - 4X^2 + 4] = E_F(X^4) - 4E_F(X^2) + 4 = 10 - 8 + 4 = 6.

This confirms D is correct: the asymptotic variance is

6

, the second moment of the influence function (i.e., its variance, since it's centered). A is wrong because it claims the variance equals

2 = E_F(X^2)

, confusing the mean of

X^2

with the variance of the influence function — these are entirely different quantities. B incorrectly equates the asymptotic variance with

E_F(X^4) = 10

, ignoring that you must subtract the squared mean of

X^2

(and cross terms) when computing the variance. C arrives at

8

by subtracting only

4E_F(X^2) = 8

from

10

, forgetting to add back the constant term

+4

in the expansion. Your study tip: always expand

E_F[(\operatorname{IF})^2]

fully using

\text{Var}(Y) = E[Y^2] - (E[Y])^2

, and verify the influence function is mean-zero before proceeding.

Question 8

For an empirical distribution $F_n$ , an estimator is well approximated by a smooth functional $T(F_n)$ . One observation $x_i$ is replaced by a new value $z$ , producing the empirical distribution $F_n^{\mathrm{rep}}$ .

Which expression gives the appropriate first-order approximation to $T(F_n^{\mathrm{rep}})-T(F_n)$ ?

$n^{-1}\{\operatorname{IF}(z;T,F_n)-\operatorname{IF}(x_i;T,F_n)\}$ , because replacement both adds and removes mass. (correct answer)
$n^{-1}\operatorname{IF}(z;T,F_n)$ , because the replacement adds contamination at the new observation.
$(n-1)^{-1}\operatorname{IF}(x_i;T,F_n)$ , because the deleted observation determines the entire first-order change.
$n^{-1}\{\operatorname{IF}(z;T,F_n)+\operatorname{IF}(x_i;T,F_n)\}$ , because both observations contribute perturbations of equal sign.

Explanation: When working with influence functions and empirical distributions, the key is tracking exactly how the distribution changes under each operation — adding mass, removing mass, or both simultaneously. When you replace observation

x_i

with a new value

z

, you are doing two things at once: removing a point mass of size

1/n

x_i

and adding a point mass of size

1/n

z

. The influence function

\operatorname{IF}(x; T, F)

captures the first-order effect of adding an infinitesimal point mass at

x

. By linearity of the first-order expansion, removing mass at

x_i

contributes

-n^{-1}\operatorname{IF}(x_i; T, F_n)

and adding mass at

z

contributes

+n^{-1}\operatorname{IF}(z; T, F_n)

, giving the net approximation

n^{-1}\{\operatorname{IF}(z;T,F_n)-\operatorname{IF}(x_i;T,F_n)\}

. This confirms A is correct. Choice B is wrong because it only accounts for the addition of

z

and completely ignores that

x_i

is simultaneously removed — a half-treatment of the perturbation. Choice C is wrong on two counts: it uses

(n-1)^{-1}

instead of

n^{-1}

, and it ignores the contribution from

z

entirely, focusing only on the deleted point. Choice D is a sign error — it adds both influence function values instead of subtracting, which would correspond to adding mass at both locations simultaneously rather than a replacement. A reliable memory anchor: replacement = addition minus deletion. Whenever a question involves swapping one observation for another, write out the two separate perturbations explicitly and track their signs before combining — this prevents both the sign error in D and the incomplete accounting in B and C.

Question 9

In a random-design linear regression model, the ordinary least-squares functional satisfies $E\{X(Y-X^{\mathsf T}\beta)\}=0$ , and $M=E(XX^{\mathsf T})$ is nonsingular. Consider contamination at a point $(x,y)$ whose residual $r=y-x^{\mathsf T}\beta$ remains fixed and nonzero while the norm of $x$ increases.

What does the influence-function calculation imply about ordinary least squares in this sequence of contamination points?

Its influence approaches zero because the fixed residual becomes negligible relative to the increasing covariate norm.
Its influence remains bounded because only the residual, rather than the covariate value, enters the estimating equation.
Its influence is generally unbounded because $M^{-1}xr$ grows with leverage when the residual is fixed. (correct answer)
Its influence is undefined because influence functions cannot be formed for random-design regression functionals.

Explanation: When analyzing robustness of an estimator, your first instinct should be to examine its influence function — the functional derivative measuring how sensitive the estimator is to infinitesimal contamination at a single point. For OLS in the random-design setting, the regression functional

\beta(F) = M^{-1}E(XY)

yields an influence function proportional to

M^{-1}x r

, where

r = y - x^\mathsf{T}\beta

is the residual at the contamination point

(x, y)

. This is exactly why C is correct. When

r

stays fixed but

\|x\| \to \infty

, the term

M^{-1}xr

grows without bound — its norm scales with

\|x\|

. This is precisely the leverage effect: a point far from the center of the covariate space exerts unbounded influence on the fitted coefficients, even with a modest residual. OLS has no mechanism to down-weight high-leverage points, so its influence function is unbounded in the covariate direction. A reverses the logic. The fixed residual doesn't "cancel out" the growing covariate norm — both enter multiplicatively in

M^{-1}xr

, and the growing norm dominates. B is factually wrong: the estimating equation

E\{X(Y - X^\mathsf{T}\beta)\} = 0

explicitly involves

X

, so the covariate absolutely enters the influence calculation. D is false — influence functions are perfectly well-defined for smooth statistical functionals, including random-design regression, as long as the functional is Gâteaux-differentiable at the model distribution. As a study strategy: whenever a question pairs "fixed residual" with "growing covariate," immediately think leverage unboundedness. This is a classic argument for why robust alternatives like M-estimators with bounded

\psi

-functions are preferred over OLS when high-leverage contamination is possible.

Question 10

Consider the variance functional $V(F)=\int (x-\mu_F)^2\,dF(x)$ . At the distribution of interest, $\mu_F=1$ and $V(F)=4$ .

What is the influence function of $V$ at a contamination point $z=4$ ?

$3$ , obtained from the contamination point's deviation from the population mean.
$5$ , obtained by centering the squared deviation by the population variance. (correct answer)
$9$ , obtained from the squared deviation of the contamination point from the mean.
$-5$ , obtained by subtracting the squared deviation from the population variance.

Explanation: When you encounter influence function questions, your goal is to quantify how a point mass contamination at

z

shifts a functional — essentially, the derivative of the functional along the direction of that contamination. For the variance functional

V(F) = \int (x - \mu_F)^2 \, dF(x)

, the influence function is derived by considering

V(F_\epsilon) = V((1-\epsilon)F + \epsilon\delta_z)

and differentiating with respect to

\epsilon

\epsilon = 0

. Working through this carefully (accounting for the fact that

\mu_F

also shifts with contamination), the influence function evaluates to

(z - \mu_F)^2 - V(F)

. Plugging in

z = 4

\mu_F = 1

, and

V(F) = 4

(4-1)^2 - 4 = 9 - 4 = 5

. This is choice B — and the description "centering the squared deviation by the population variance" captures precisely this structure. Choice A gives

3

, which is just

z - \mu_F = 4 - 1

, the raw deviation — not squared, not centered. This ignores almost all of the influence function's structure. Choice C gives

9 = (z - \mu_F)^2

, which is the squared deviation alone; this student forgot to subtract

V(F)

, missing the centering step entirely. Choice D gives

-5

, which reverses the subtraction — subtracting the squared deviation from the variance rather than the variance from the squared deviation. A useful memory device: the influence function of variance looks like a "centered" squared deviation —

(z - \mu)^2 - \sigma^2

— analogous to how variance itself centers squared deviations around their mean. Always remember to subtract

V(F)