Statistics Graduate Level Quiz: Kernel Density Estimation

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Kernel Density EstimationQuestion 1 of 10

For observations $0,1,2$ , consider the kernel density estimator $\widehat f_h(x)=\frac{1}{nh}\sum_{i=1}^n K\!\left(\frac{x-X_i}{h}\right),$ where $h=\frac12$ and $K(u)=\frac{15}{16}(1-u^2)^2\mathbf 1\{|u|\leq 1\}.$ What is $\widehat f_h\!\left(\frac34\right)$ ?

\frac{45}{128}

\frac{135}{256}

\frac{45}{256}

\frac{135}{128}

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Kernel Density Estimation

Practice Kernel Density Estimation 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 Kernel Density Estimation, 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

$\frac{45}{128}$ (correct answer)
$\frac{135}{256}$
$\frac{45}{256}$
$\frac{135}{128}$

Explanation: Kernel density estimation questions test your ability to mechanically apply the KDE formula, which requires careful bookkeeping of which kernels actually "activate" at your target point. At

x = \frac{3}{4}

with

h = \frac{1}{2}

and

n = 3

, the estimator is:

\widehat{f}_h\!\left(\tfrac{3}{4}\right) = \frac{1}{3 \cdot \frac{1}{2}}\sum_{i=1}^{3} K\!\left(\frac{\frac{3}{4} - X_i}{\frac{1}{2}}\right) = \frac{2}{3}\sum_{i=1}^{3} K\!\left(\frac{\frac{3}{4} - X_i}{\frac{1}{2}}\right)

The Epanechnikov kernel

K(u) = \frac{15}{16}(1-u^2)^2

is zero when

|u| > 1

. Check each observation:

$X_i = 0$ : $u = \frac{3/4}{1/2} = \frac{3}{2}$ , so $|u| > 1$ — kernel is zero.
$X_i = 1$ : $u = \frac{-1/4}{1/2} = -\frac{1}{2}$ , so $K(-\tfrac{1}{2}) = \frac{15}{16}(1 - \tfrac{1}{4})^2 = \frac{15}{16} \cdot \frac{9}{16} = \frac{135}{256}$ .
$X_i = 2$ : $u = \frac{-5/4}{1/2} = -\frac{5}{2}$ , so $|u| > 1$ — kernel is zero.

Only

X_i = 1

contributes, giving:

\widehat{f}_h\!\left(\tfrac{3}{4}\right) = \frac{2}{3} \cdot \frac{135}{256} = \frac{270}{768} = \frac{45}{128}

That confirms A. Choice B,

\frac{135}{256}

, is the trap of forgetting to multiply by the prefactor

\frac{2}{3}

— you evaluated the kernel but skipped the normalization. Choice D,

\frac{135}{128}

, doubles instead of multiplying by

\frac{2}{3}

. Choice C,

\frac{45}{256}

, uses

\frac{1}{3}

as the prefactor instead of

\frac{2}{3}

, confusing

\frac{1}{n}

with

\frac{1}{nh}

. Always compute the standardized argument

u = (x - X_i)/h

first to identify which observations fall within the kernel's support before evaluating anything else.

Question 2

For a twice-differentiable density and a symmetric second-order kernel, suppose the leading asymptotic mean integrated squared error has the form $\operatorname{AMISE}(h)=C_1h^4+\frac{C_2}{nh},$ where $C_1$ and $C_2$ are positive constants. Restrict attention to bandwidths of the form $h_n=n^{-\alpha}.$ Which choice yields the fastest decay of the displayed AMISE while also satisfying the usual consistency conditions?

$\alpha=0,$ because a fixed bandwidth keeps the variance of the estimator asymptotically negligible.
$\alpha=\frac15,$ because it balances the integrated squared-bias and variance terms. (correct answer)
$\alpha=\frac13,$ because it makes the bandwidth shrink more rapidly without violating consistency.
$\alpha=1,$ because it eliminates smoothing bias at the fastest asymptotically possible rate.

Explanation: Kernel density estimation questions hinge on the bias-variance tradeoff baked into the AMISE. When you see a form like

C_1 h^4 + \frac{C_2}{nh}

, your instinct should be: minimize over

h

, then check what power of

n

that optimal bandwidth takes. Taking the derivative with respect to

h

and setting it to zero gives

4C_1 h^3 - \frac{C_2}{nh^2} = 0,

which yields

h^* \propto n^{-1/5}.

So the optimal bandwidth has the form

h_n = n^{-\alpha}

with

\alpha = \frac{1}{5}.

Plugging back in, both terms contribute equally and AMISE decays at rate

n^{-4/5}

— the fastest possible for this setup. Consistency also requires

h \to 0

and

nh \to \infty

, both satisfied since

\alpha = 1/5

means

h \to 0

and

nh = n^{4/5} \to \infty

. That confirms B is correct. Choice A (

\alpha = 0

) fixes

h

as a constant, so

h \not\to 0

and the bias never vanishes — the estimator is inconsistent. Choice C (

\alpha = 1/3

) makes

h

shrink faster than optimal, so the variance term

C_2/(nh) \propto n^{-2/3}

dominates and AMISE decays only at rate

n^{-2/3}

, which is slower than

n^{-4/5}

. It's consistent but suboptimal. Choice D (

\alpha = 1

) shrinks

h

too aggressively:

nh = n^0 = 1

, so

nh \not\to \infty

and the variance explodes — another inconsistent choice. Study tip: Always verify two things for KDE bandwidth sequences: the AMISE-minimizing rate (set derivative to zero) and the consistency conditions (

h \to 0

nh \to \infty

). Forgetting the second step is the most common trap on these questions.

Question 3

Independent observations are sampled from the uniform density on $[0,1].$ A conventional kernel density estimator is used with a symmetric kernel satisfying $\int K(u)\,du=1.$ No boundary correction is applied, and the bandwidth sequence satisfies $h_n\to0$ and $nh_n\to\infty.$

What is the probability limit of the estimator evaluated exactly at the left endpoint, $\widehat f_{h_n}(0)$ ?

It converges to $0$ because kernels centered on positive observations contribute no limiting mass at the endpoint.
It converges to $\frac12$ because only one half of the symmetric kernel mass overlaps the support. (correct answer)
It converges to $1$ because consistency holds wherever the underlying density is positive.
It converges to $2$ because the kernel mass inside the support is implicitly renormalized.

Explanation: Kernel density estimation at boundary points is a classic source of asymptotic bias, and recognizing why requires you to think carefully about how mass integrates over a truncated region. The kernel density estimator is

\widehat{f}_{h_n}(x) = \frac{1}{nh_n}\sum_{i=1}^n K\!\left(\frac{x - X_i}{h_n}\right).

Its pointwise probability limit equals its limiting expected value (by the law of large numbers under standard bandwidth conditions). At any interior point,

E[\widehat{f}_{h_n}(x)] \to f(x)

because the full kernel integrates to 1 over the support. But at

x = 0

, the situation changes: the kernel is centered at the boundary, and since the data live only on

[0,1]

, only the right half of the symmetric kernel actually overlaps the support. Formally,

E[\widehat{f}_{h_n}(0)] = \int_0^\infty \frac{1}{h_n} K\!\left(\frac{-t}{h_n}\right) f(t)\,dt \;\to\; f(0)\int_0^\infty K(u)\,du = 1 \cdot \tfrac{1}{2} = \tfrac{1}{2}.

Because

f \equiv 1

[0,1]

, the limit is

\frac{1}{2}

, confirming answer B. Choice A is wrong because contributions from nearby positive observations do not vanish — they simply get cut in half. Choice C applies interior consistency results where boundary effects are absent; consistency fails at the boundary precisely because the estimator carries systematic bias toward zero. Choice D invents a renormalization that a standard kernel estimator never performs — that would describe a boundary-corrected estimator. As a study tip: whenever a boundary is present, immediately check which fraction of kernel mass lies inside the support. Symmetric kernels lose exactly half their mass at a hard boundary, so the limiting bias factor is always

\frac{1}{2}

Question 4

A kernel with support $[-1,1]$ is used with a fixed bandwidth $h$ and a sample of size $n.$ At a particular point $x$ , the current estimate $\widehat f_h(x)$ is positive. One additional observation is added, but its distance from $x$ exceeds $h.$ The bandwidth is not recomputed.

How does the kernel density estimate at $x$ change after adding the observation?

It remains $\widehat f_h(x)$ because the new observation contributes zero at $x$ .
It becomes $\widehat f_h(x)+\frac{1}{(n+1)h}$ because every added observation contributes unit kernel mass.
It becomes $\frac{n+1}{n}\widehat f_h(x)$ because the effective sample size has increased.
It becomes $\frac{n}{n+1}\widehat f_h(x)$ because only the normalization changes at $x$ . (correct answer)

Explanation: Whenever you see a question about kernel density estimation, anchor yourself to the formula:

\widehat{f}_h(x) = \frac{1}{nh} \sum_{i=1}^{n} K\!\left(\frac{x - X_i}{h}\right)

Every observation contributes a scaled kernel bump, but only observations within bandwidth

h

x

contribute nonzero values when the kernel has compact support

[-1,1]

. Here's the key insight: the new observation is farther than

h

from

x

, so

K\!\left(\frac{x - X_{n+1}}{h}\right) = 0

. Its numerator contribution is exactly zero. However, the denominator still grows — the sample size increases from

n

n+1

. The updated estimate becomes:

\widehat{f}_{h,\text{new}}(x) = \frac{1}{(n+1)h} \sum_{i=1}^{n} K\!\left(\frac{x - X_i}{h}\right) = \frac{n}{n+1} \cdot \widehat{f}_h(x)

So the estimate at

x

is diluted by the factor

\frac{n}{n+1}

, making D correct. A is tempting but incomplete — yes, the new observation contributes zero kernel mass, but you cannot ignore how the normalization constant changes. B is wrong because it assumes the new observation contributes a full

\frac{1}{(n+1)h}

bump, which requires

K(\cdot) = 1

, impossible when the observation falls outside the kernel's support. C inverts the logic entirely — a larger sample size dilutes the existing contributions rather than inflating them. The study tip: always track both the numerator (kernel evaluations) and the denominator (

nh

) separately. In KDE, adding a remote observation affects only the denominator, quietly shrinking every existing estimate.

Question 5

A bivariate product-kernel density estimator uses the same scalar bandwidth $h$ in both coordinates. Under standard smoothness assumptions, its leading asymptotic mean integrated squared error has order $h^4+(nh^2)^{-1}.$

Which bandwidth order and resulting optimized error order follow from balancing these terms?

$h\asymp n^{-1/5}$ and $\operatorname{AMISE}\asymp n^{-4/5},$ as in the univariate case.
$h\asymp n^{-1/3}$ and $\operatorname{AMISE}\asymp n^{-2/3},$ because variance depends on the squared bandwidth.
$h\asymp n^{-1/4}$ and $\operatorname{AMISE}\asymp n^{-1/2},$ because two coordinates double the bias order.
$h\asymp n^{-1/6}$ and $\operatorname{AMISE}\asymp n^{-2/3},$ reflecting the two-dimensional variance term. (correct answer)

Explanation: Whenever you see a kernel density estimation question involving AMISE balancing, your job is to treat bias and variance as two terms in an equation and find the bandwidth that minimizes their sum — then substitute back to find the optimal error rate. Here, the AMISE has two terms: a bias-squared term

h^4

and a variance term

(nh^2)^{-1}

. To find the optimal

h

, set them equal in order of magnitude:

h^4 \asymp \frac{1}{nh^2}

Multiply both sides by

nh^2

nh^6 \asymp 1 \implies h \asymp n^{-1/6}

Substituting back, the bias term gives

(n^{-1/6})^4 = n^{-2/3}

, and the variance term gives

(n \cdot n^{-2/6})^{-1} = n^{-2/3}

. Both terms are

n^{-2/3}

, confirming

\operatorname{AMISE} \asymp n^{-2/3}

. This is answer D. A is the classic univariate result, where the variance term is

(nh)^{-1}

. In 2D with a product kernel, the variance inflates to

(nh^2)^{-1}

because the effective local bin volume scales as

h^2

, not

h

. Plugging in univariate logic here gives the wrong exponent. B arrives at the right AMISE rate but the wrong bandwidth. It appears to account for the squared bandwidth in the variance but mis-solves the balance equation — setting

h^4 \asymp (nh^2)^{-1}

does not yield

h \asymp n^{-1/3}

. C incorrectly supposes that adding a second coordinate changes the bias order, but the bias remains

O(h^2)

per dimension with a squared-kernel product, keeping the squared bias at

h^4

. Study tip: In

d

dimensions, the product-kernel variance term is

(nh^d)^{-1}

, so the optimal bandwidth scales as

n^{-1/(4+d)}

and AMISE as

n^{-4/(4+d)}

. For

d=2

n^{-1/6}

and

n^{-2/3}

. Memorizing this general formula saves time on any dimensionality question.

Question 6

A researcher converts measurements by defining $Y=3X+2.$ She computes a kernel density estimate from the transformed observations using the same base kernel as for the original data.

If the original bandwidth is $h_X$ , which bandwidth and density relationship make the kernel density estimator exactly equivariant under this transformation?

Use $h_Y=h_X$ , giving $\widehat f_Y(y)=\frac13\widehat f_X\!\left(\frac{y-2}{3}\right).$
Use $h_Y=3h_X$ , giving $\widehat f_Y(y)=3\widehat f_X\!\left(\frac{y-2}{3}\right).$
Use $h_Y=3h_X$ , giving $\widehat f_Y(y)=\frac13\widehat f_X\!\left(\frac{y-2}{3}\right).$ (correct answer)
Use $h_Y=h_X/3$ , giving $\widehat f_Y(y)=\widehat f_X\!\left(\frac{y-2}{3}\right).$

Explanation: Whenever you see a question about kernel density estimation (KDE) under linear transformations, anchor your thinking to two requirements simultaneously: how the bandwidth must scale, and how the density value must transform to preserve proper probability mass. For a linear transformation

Y = aX + b

with

a > 0

, the KDE achieves exact equivariance — meaning it behaves like the true density under the transformation — when the bandwidth scales as

h_Y = a \cdot h_X

. Here,

a = 3

, so

h_Y = 3h_X

. This makes intuitive sense: stretching the data by a factor of 3 spreads observations apart, so the smoothing window must widen proportionally to maintain the same relative smoothing. For the density relationship, recall that if

Y = 3X + 2

, then the change-of-variables formula requires

f_Y(y) = \frac{1}{|a|} f_X\!\left(\frac{y-b}{a}\right) = \frac{1}{3} f_X\!\left(\frac{y-2}{3}\right)

. The factor

\frac{1}{3}

ensures the density integrates to 1 after stretching. Combining both requirements confirms C is correct. A uses the wrong bandwidth (

h_Y = h_X

instead of

3h_X

) but accidentally gets the density formula right — it's a half-correct answer that fails equivariance because the bandwidth doesn't account for the scale change. B has the correct bandwidth

h_Y = 3h_X

but inverts the density scaling factor, writing

3\widehat{f}_X

instead of

\frac{1}{3}\widehat{f}_X

. This would produce a density that doesn't integrate to 1 — a fundamental violation. D shrinks the bandwidth (

h_Y = h_X/3

) when it should expand, and omits the necessary

\frac{1}{3}

normalizing factor entirely. A reliable memory aid: for

Y = aX + b

, bandwidth multiplies by

|a|

and the density divides by

|a|

. These two adjustments always move in opposite directions.

Question 7

A continuous sample contains no tied observations. A researcher proposes choosing the bandwidth by maximizing the ordinary in-sample log-likelihood $\sum_{i=1}^n\log \widehat f_h(X_i)$ rather than the leave-one-out log-likelihood $\sum_{i=1}^n\log \widehat f_{h,-i}(X_i).$ The kernel is Gaussian.

Why is the ordinary in-sample criterion unsuitable for bandwidth selection in this setting?

It tends to favor extremely small bandwidths because each observation contributes a large kernel value to its own fitted density. (correct answer)
It tends to favor extremely large bandwidths because each observation is omitted from its own fitted density.
It is independent of the bandwidth because every Gaussian kernel integrates to one over the real line.
It is unbiased for predictive log-likelihood but has larger variance than the leave-one-out criterion.

Explanation: When choosing a bandwidth for kernel density estimation, you need a criterion that genuinely measures how well your estimated density generalizes — not just how well it memorizes the data. The in-sample log-likelihood

\sum_{i=1}^n \log \widehat{f}_h(X_i)

is fatally flawed because each observation

X_i

participates in its own estimated density value

\widehat{f}_h(X_i)

. With a Gaussian kernel, the contribution of observation

X_i

to its own density estimate is

\frac{1}{nh}\phi(0) = \frac{1}{nh\sqrt{2\pi}},

which grows without bound as

h \to 0

. As the bandwidth shrinks, each point's self-contribution dominates and inflates the log-likelihood artificially. The optimizer therefore drives

h

toward zero, producing degenerate "spike" estimates — one sharp Gaussian centered at each data point — rather than a smooth, meaningful density. This confirms that A is correct. B is wrong because it describes the leave-one-out criterion, not the in-sample one. In leave-one-out cross-validation,

X_i

is deliberately excluded from

\widehat{f}_{h,-i}(X_i)

, which is exactly why that approach avoids the collapse-to-zero problem. C is wrong because the criterion absolutely depends on

h

. While each Gaussian kernel integrates to one (a property ensuring

\widehat{f}_h

is a valid density), the pointwise evaluation at each

X_i

changes with

h

— the two concepts should not be conflated. D is wrong because the in-sample criterion is not an unbiased estimator of predictive log-likelihood; it is systematically upward-biased due to the self-evaluation problem described above. As a study tip: whenever a selection criterion evaluates a model on the same data used to fit it, ask whether any observation is "cheating" by inflating its own fit — this self-evaluation trap appears frequently in density estimation and model selection questions.

Question 8

At an interior point $x_0$ , a density satisfies $f''(x_0)=-6.$ A symmetric kernel has second moment $\mu_2(K)=1,$ and the bandwidth is $h=0.1.$ Terms of order smaller than $h^2$ may be ignored.

Which approximation to the expectation of the kernel density estimator at $x_0$ is correct?

$\operatorname{E}[\widehat f_h(x_0)]\approx f(x_0)+0.03,$ reflecting upward smoothing bias at a local maximum.
$\operatorname{E}[\widehat f_h(x_0)]\approx f(x_0)-0.03,$ reflecting downward smoothing bias at a local maximum. (correct answer)
$\operatorname{E}[\widehat f_h(x_0)]\approx f(x_0)-0.06,$ because the leading bias equals $h^2f''(x_0).$
$\operatorname{E}[\widehat f_h(x_0)]\approx f(x_0)-0.30,$ because the leading bias is linear in the bandwidth.

Explanation: Whenever you see a question about kernel density estimation bias, your first instinct should be to recall the standard bias expansion. For a symmetric kernel at an interior point, the bias formula is:

\operatorname{E}[\widehat{f}_h(x_0)] \approx f(x_0) + \frac{1}{2}h^2\mu_2(K)f''(x_0)

This result comes from a Taylor expansion of

f

inside the convolution integral; the symmetry of the kernel kills the odd-order terms, leaving the leading bias proportional to

h^2 f''(x_0)

. Plugging in the given values —

h = 0.1

\mu_2(K) = 1

, and

f''(x_0) = -6

— you get:

\text{Bias} = \frac{1}{2}(0.1)^2(1)(-6) = \frac{1}{2}(0.01)(-6) = -0.03

\operatorname{E}[\widehat{f}_h(x_0)] \approx f(x_0) - 0.03

, which is answer B. The negative sign makes intuitive sense: a point where

f''(x_0) < 0

is a local maximum, and smoothing pulls estimates downward toward neighboring lower values — a classic downward bias. A gets the sign wrong. Upward bias would require

f'' > 0

(a local minimum), not a local maximum. C omits the crucial

\frac{1}{2}

factor and uses

h^2 f''(x_0) = -0.06

directly — a common algebraic error that forgets the coefficient in the Taylor expansion. D claims the bias is linear in

h

, which contradicts the theory; the leading bias is always of order

h^2

for symmetric kernels. The key memory anchor: bias

= \frac{1}{2}h^2\mu_2(K)f''(x_0)

. Never drop the

\frac{1}{2}

, and always check the sign of

f''

to determine the direction of smoothing bias.

Question 9

Two symmetric second-order kernels, $K_1$ and $K_2$ , both integrate to one and have the same second moment. They satisfy $R(K_1)<R(K_2),$ where $R(K)=\int K(u)^2\,du.$ The same bandwidth sequence is used with each kernel to estimate a twice-differentiable density at an interior point.

Which comparison is correct to leading asymptotic order?

$K_1$ has the same leading bias and a smaller leading variance than $K_2$ . (correct answer)
$K_1$ has a smaller leading bias and the same leading variance as $K_2$ .
$K_1$ has both a larger leading bias and a smaller leading variance than $K_2$ .
$K_1$ has the same leading bias and a larger leading variance than $K_2$ .

Explanation: Kernel density estimation questions like this one test whether you understand which kernel properties govern bias versus variance in the asymptotic MSE decomposition. For a kernel estimator with bandwidth

h \to 0

, the leading bias at an interior point depends on

\mu_2(K) = \int u^2 K(u)\,du

— the second moment of the kernel — through the expression

\frac{h^2}{2}\mu_2(K) f''(x)

. The leading variance depends on

R(K) = \int K(u)^2\,du

through

\frac{1}{nh}R(K)f(x)

. These two quantities are entirely separate: bias is controlled by the second moment, variance by the roughness functional

R(K)

. The passage tells you that

K_1

and

K_2

share the same second moment but differ in roughness, with

R(K_1) < R(K_2)

. Since both kernels are symmetric, integrate to one, and have identical second moments, their leading bias terms are identical. However, because

R(K_1) < R(K_2)

K_1

produces a strictly smaller leading variance. This confirms A as correct. Choice B is wrong because it claims

K_1

has a smaller bias — but bias depends only on the second moment, which is equal for both kernels. Choice C compounds this error by also claiming a larger bias for

K_1

, which contradicts the identical second moments. Choice D gets the variance comparison exactly backwards: smaller

R(K)

means smaller variance, so

K_1

's advantage is lower variance, not higher. A useful rule to memorize: bias tracks the second moment; variance tracks

R(K)

. These are independent kernel properties, so a question changing one while fixing the other affects only one component of MSE.

Question 10

A researcher uses a signed higher-order kernel satisfying $\int K(u)\,du=1,$ but $K(u)<0$ on part of its support. The estimator retains the usual form $\widehat f_h(x)=\frac{1}{nh}\sum_{i=1}^nK\!\left(\frac{x-X_i}{h}\right).$

Which statement about the resulting estimator is necessarily correct?

It is nonnegative everywhere but may fail to integrate to one because the kernel has negative lobes.
It integrates to less than one whenever at least one observation lies near a negative kernel lobe.
It integrates to one but can be negative at some points, so it need not be a genuine density. (correct answer)
It is a genuine density whenever the sample size is sufficiently large, regardless of the realized sample.

Explanation: When you encounter a question about higher-order or signed kernels, the key is carefully separating two properties that a genuine probability density must satisfy: nonnegativity everywhere, and integrating to one. These two properties are logically independent, and signed kernels break them asymmetrically. The condition

\int K(u)\,du = 1

is preserved under the estimator's construction. Since

\widehat{f}_h(x) = \frac{1}{nh}\sum_{i=1}^n K\!\left(\frac{x-X_i}{h}\right)

, integrating over all

x

gives

\int \widehat{f}_h(x)\,dx = \frac{1}{n}\sum_{i=1}^n \int K\!\left(\frac{x-X_i}{h}\right)\frac{dx}{h} = \frac{1}{n}\sum_{i=1}^n 1 = 1,

regardless of the sample. So the estimator always integrates to one. However, where

K(u) < 0

, the estimator can take negative values for certain

x

, violating nonnegativity. This makes C correct: the estimator integrates to one but is not guaranteed to be a valid density. Answer A is wrong because the estimator is not necessarily nonnegative — negative kernel lobes directly produce negative estimates. It also incorrectly claims the integral deviates from one, which contradicts the linearity argument above. Answer B is wrong for the same integral reason: the integral is always exactly one, not "less than one," because each kernel shift integrates to one regardless of where observations fall. Answer D is wrong because no sample size, however large, eliminates the possibility of negative values at specific points — that is a structural feature of the signed kernel, not a finite-sample artifact. Your takeaway: for signed kernels, the normalization condition travels through linearity cleanly, but nonnegativity does not. Always ask "which density property does linearity preserve?" — the answer is integration to one, never pointwise positivity.