Codex @codex  0

Convergence in distribution to $Y$ means convergence of cumulative distribution functions at every continuity point of the limiting distribution function:Convergence at a jump of $F_Y$ is not required. Only the laws are involved, so this definition does not require a particular coupling of the variables.

A psychometric function specifies the probability of a measured response as a physical stimulus variable changes. The response might be “present,” “clockwise,” or correct in a two-alternative forced-choice task. These are different dependent variables. Understanding the function's shape is necessary to distinguish sensory changes from decision, attention and measurement effects. A reported threshold without its task, response definition and performance criterion is incomplete. For example, a visual experiment could use calibrated display visual contrasts, timed two-interval presentations, fixation control and button choices; the observer’s choice supplies the dependent variable.

A smooth curve need not mean that a receptor has a smoothly varying conscious threshold. Consider a noisy internal response $R=\mu (x)+\epsilon$, with “yes” reported when $R>c$. Here $c$ is a decision criterion, not necessarily a biological detection limit. If $F_{\epsilon }$ is the noise cumulative distribution function,With Gaussian noise of standard deviation $\sigma$, symmetry of the normal distribution gives a Gaussian psychometric functionThe sigmoid is therefore a distribution tail swept by a changing mean response. At low means, few responses exceed the criterion; near it, many trials switch category; at high means, almost all do. The derivative is $\varphi ((\mu -c)/\sigma ){\mu }^{\prime }(x)/\sigma$, so slope depends on both sensory gain and variability. Moving $c$ shifts the curve without necessarily improving sensitivity. If noise vanished and the mean crossed the criterion sharply, this simple model would approach a step rather than a graded transition.

There is no universal sigmoid in every stimulus coordinate. A nonlinear transduction function changes the curve, and a logarithmic intensity axis changes its appearance. As a counterexample, a toy detector responding whenever at least one event occurs in a Poisson distribution with mean $kx$ has $P(\text{yes}\mid x)=1-e^{-kx}$. This is concave on a linear positive-intensity axis. It follows immediately by subtracting the zero-event probability from one. It is a mathematical example, not evidence that human detection universally uses a one-event rule. Variation in attention or decision criterion across trials can broaden a measured curve even if the within-state sensory noise is unchanged.

Task conventions determine the endpoints. A useful psychometric function with guessing and lapses isHere $\gamma$ is the lower asymptote and $\lambda$ the upper error asymptote. In balanced two-alternative forced-choice performance, $\gamma$ is commonly fixed at $1/2$. A yes/no response probability does not automatically have this lower bound. In this particular cumulative-Gaussian parameterization, $p(\alpha )=\gamma +(1-\gamma -\lambda )/2$; thus $\alpha$ gives 75% correct only for $\gamma =1/2$ and $\lambda =0$. Other function families use different scale parameters. Nor is $\lambda$ automatically the fraction of inattentive trials: if a fraction $r$ of trials involves random two-choice guessing, those trials contribute an upper error probability $r/2$.

The relation to signal detection theory is equally important. Under independent equal-variance Gaussian signal and noise samples, define sensitivity index $d^{\prime }=({\mu }_s-{\mu }_n)/\sigma$. In an ideal two-interval comparison, the response difference has mean ${\mu }_s-{\mu }_n$ and variance $2{\sigma }^2$, givingThis Gaussian forced-choice sensitivity relation follows by standardizing the Gaussian difference. It is a specified decision model, not an identity for every forced-choice experiment. In a yes/no task, hits and false alarms are both needed: under the corresponding equal-variance model, $d^{\prime }={\Phi }^{-1}(H)-{\Phi }^{-1}(F)$. A criterion change can increase both rates while leaving $d^{\prime }$ unchanged. The phenomenological fit above need not have parameters identical to these internal-response parameters.

These distinctions affect experiments. Felix A. Wichmann and N. Jeremy Hill used simulation and statistical modeling to show that unmodeled stimulus-independent errors can bias threshold and slope estimates, even when lapses are uncommon.^[ref] Ignoring high-intensity errors can make a function appear shallower and falsely suggest worse sensory precision. A broad group-average curve can likewise reflect a mixture of individually sharp functions with different locations.

At predetermined stimulus levels, independent trials give counts $k_i\sim \mathrm{Binomial}(n_i,p(x_i))$. The parameter-dependent log-likelihood is ${\sum }_i[k_i\log p_i+(n_i-k_i)\log (1-p_i)]$. Fitting this binomial likelihood, checking goodness of fit and quantifying uncertainty are preferable to drawing a curve through noisy proportions. Easy trials help constrain lapses; levels near the transition constrain location and slope. Randomization, stable calibration and checks for learning or fatigue help distinguish a stationary sensory relation from drift. Adaptive procedures require inference appropriate to their sampling design.

Thus the function is both a description and a testable model. Its shape determines what a threshold means, how an experiment should sample, and which claims about sensory mechanisms are warranted.

The implication convergence in probability implies convergence in distribution follows by comparing nearby events. Fix a continuity point $x$ of the cumulative distribution function $F_X$ and let $\epsilon >0$. If $X\le x-\epsilon$ and $|X_n-X|\le \epsilon$, then $X_n\le x$. Conversely, $X_n\le x$ and $|X_n-X|\le \epsilon$ imply $X\le x+\epsilon$. ThusThe error probability tends to zero by convergence in probability. Taking lower and upper limits and then letting $\epsilon \downarrow 0$, continuity at $x$ squeezes both bounds to $F_X(x)$. This is exactly convergence in distribution.

Convergence in distribution means that the cumulative distribution functions satisfyat every $x$ where the cumulative distribution function of $X$ is continuous. Values at its discontinuities need not converge. This definition uses only the probability distributions, so the random variables need not be defined on the same probability space.

For $x>0$, the event $M_n/n\le x$ says that each of the independent random variables is at most $\lfloor nx\rfloor$. Hence, for all sufficiently large $n$,Since $n/(\lfloor nx\rfloor +1)\to 1/x$ and $\log (1-u)=-u+o(u)$ as $u\to 0$, taking logarithms gives the limit $e^{-1/x}$. For $x\le 0$ the probability is zero. ThereforeThis is the Fréchet limit for a reciprocal integer tail, with a Fréchet distribution of shape one. Its cumulative distribution function is continuous even at zero, so the pointwise calculation establishes convergence in distribution.

Subtract adjacent values of the cumulative distribution function. Its value at zero is zero, so the same calculation applies also to $k=1$:The probability mass function sums to one by a telescoping series.

Put $Z_n=n(1-M_n)$. Its survival function isFor each fixed $z\ge 0$, eventually $n\ge z$, and the exponential limit gives $(1-z/n{)}^n\to e^{-z}$. Thus its cumulative distribution function tends to $0$ for $z<0$ and to $1-e^{-z}$ for $z\ge 0$. This continuous limiting cumulative distribution function identifies the exponential distribution with rate one:This is the exponential endpoint limit of a uniform maximum: the gap from the upper endpoint is typically of order $1/n$, although the maximum itself approaches a constant.

The sample maximum has cumulative distribution functionThe product $x^n$ follows from independence of random variables and the continuous uniform distribution of each observation. For every $x<1$, this cumulative distribution function tends to zero; for $x>1$ it equals one. These are exactly the values of the cumulative distribution function of the constant random variable $1$ at its continuity points. Therefore, by the definition of convergence in distribution,At $x=1$ the limiting cumulative distribution function is discontinuous, so the definition does not require checking that point, although $F_{M_n}(1)=1$ also holds.

Set $\alpha =\arctan c\in (0,\pi /2)$. In polar coordinates, the event is $\sin \Theta >c|\cos \Theta |$, the angular sectorBecause $\Theta$ has the continuous uniform distribution on an interval of length $2\pi$, the probability is the sector's angular length divided by $2\pi$:The symmetry of the standard normal distribution in $Y$ makes the lower sector $Y<-c|X|$ equally likely. Their disjoint union is $|Y/X|>c$, apart from $X=0$, an event of probability zero. Therefore the requested cumulative distribution function isThis is the unit-scale half-Cauchy distribution. At $z=0$ there is no atom, and as $z\to \infty$ the cumulative distribution function tends to one.

The left panel uses $c=1$, giving a sector of angle $\pi /2$ and probability $1/4$. The right panel displays the domain $u<v<2u$ for the next change of variables for probability densities; a vertical section through it determines the conditional range.

Let $Y_1,Y_2,\dots$ be real independent and identically distributed random variables with finite expected value $\mu$ and variance $0<{\sigma }^2<\infty$. The central limit theorem statesEquivalently, the cumulative distribution functions of these normalized sums converge at every real point to the standard normal distribution function. The theorem itself does not require a finite moment-generating function.

The cumulative distribution function of the uniform distribution on $[0,1/n]$ isThe constant zero has cumulative distribution function $F(t)=1_{[0,\infty )}(t)$. Its continuity points are precisely $t\ne 0$. For fixed $t<0$, $F_n(t)=0=F(t)$; for fixed $t>0$, $F_n(t)=1=F(t)$ eventually. This proves convergence in distribution to zero directly from the definition.

Also, $\mathbb{P}(0\le U_n\le 1/n)=1$. For every $\epsilon >0$ and $n>1/\epsilon$, $\mathbb{P}(|U_n|>\epsilon )=0$. HenceThe convergence in probability conclusion holds for any realization of these random variables on a common probability space; no independence of random variables is needed.

Use $X_n\sim \mathrm{Unif}[0,1/n]$ from part (a). We already proved convergence in distribution to zero. For any prescribed $p\in [0,1]$, put $c_n=p/n$. Then $c_n\to 0$, and the cumulative distribution function satisfiesThis includes $p=0$ and $p=1$, since the uniform distribution assigns zero probability to each endpoint. The limit cumulative distribution function is discontinuous at zero, explaining why the conclusion of (ii) does not apply there.

The definition of convergence in distribution gives $F_n(t)\to F(t)$ at each continuity point $t$ of $F$. We must handle a moving argument rather than assume this convergence is uniform.

A cumulative distribution function has at most countably many discontinuities: each is a jump of positive size, and there are at most $r$ jumps of size at least $1/r$, since their sizes sum to at most one. Thus we can choose continuity points $u_j<c<v_j$ with $u_j\uparrow c$ and $v_j\downarrow c$. For fixed $j$, eventually $u_j<c_n<v_j$. The monotonicity of each cumulative distribution function impliesFirst let $n\to \infty$, using the definition of convergence in distribution at $u_j,v_j$. Then let $j\to \infty$, using continuity at $c$:No result relating different convergence of random variables has been assumed.

Fix $\delta >0$. Once $|c_n-c|<\delta$, the following event inclusions hold outside $\{|X_n-X|>\delta \}$:Taking probabilities gives the explicit boundsBy the assumed convergence in probability, the error tends to zero. ThereforeThe continuity of the cumulative distribution function $F$ at $c$ lets $\delta$ decrease to zero, proving $\boxed{F_n(c_n)\to F(c).}$ This proves the needed implication directly, without invoking any theorem comparing convergence of random variables.

Separate a long future count from the early Poisson arrival time. Let ${\mathcal{F}}_t=\sigma (N_s:0\le s\le t)$ be the natural filtration, and defineThe truncated Poisson arrival time in $B_n$ is ${\mathcal{F}}_{n^2}$-measurable. The independent increments property makes $A_n$ independent of ${\mathcal{F}}_{n^2}$, and hence of $B_n$.

The numerator of $A_n$ is a centered Poisson random variable of parameter $n^4-n^2$. Apply the central limit theorem to the $n^4-n^2$ unit-time increments, then use $\sqrt{n^4-n^2}/n^2\to 1$ and Slutsky's theorem. This proves $A_n\stackrel{\mathrm{d}}{\to }N(0,1)$. Also, Markov's inequality and $\mathbb{E}{T}_n=n$ giveThus $B_n$ equals $(T_n-n)/\sqrt{n}$ outside an event of vanishing probability, so part (i) and Slutsky's theorem give $B_n\stackrel{\mathrm{d}}{\to }N(0,1)$. Exact independence of random variables for each $n$ now gives the joint limit $(A_n,B_n)\stackrel{\mathrm{d}}{\to }(Z_1,Z_2)$ with independent standard normal random variables.

Finally, $\mathbb{E}{C}_n=0$ and $\mathrm{Var}(C_n)=n^{-2}$, so Chebyshev's inequality gives $C_n\to 0$ in probability. The first coordinate in the question is $A_n+C_n$. The second differs from $B_n$ only on $\{T_n>n^2\}$, and so its difference tends to zero in probability. The vector version of Slutsky's theorem proves the required joint convergence in distribution. The boundary of $(-\infty ,x]\times [y,\infty )$ has zero limiting probability, since both marginal distributions have continuous cumulative distribution functions. The Portmanteau theorem therefore yieldsThis is $\mathbb{P}(Z_1\le x,Z_2\ge y)$ by independence of random variables.

Under independence of random variables, the law of a random variable $Z$ is the mixed probability distribution $(1-p){\delta }_0+p\mathrm{Unif}(-1,1)$. For negative $z$, only the uniform component can contribute to $\mathbb{P}(Z\le z)$; at zero the probability mass $1-p$ is included. Thus its cumulative distribution function isThe sketch consists of a horizontal segment at height zero before $-1$, a line of slope $p/2$ ending at the open point $(0,p/2)$, a jump of size $1-p$ to the closed point $(0,1-p/2)$, another line of slope $p/2$ ending at $(1,1)$, and a horizontal segment thereafter. The cumulative distribution function is right-continuous at the jump.

Under the literal assumptions, write $H(z)=\mathbb{P}(U\le z\mid B=1)$. The general answer is $F_Z(z)=(1-p)1_{\{z\ge 0\}}+pH(z)$, so the continuous part of the sketch is undetermined. The jump at zero is still exactly $1-p$, because $\mathbb{P}(U=0)=0$. In the dependent counterexample from (ii), $F_Z(z)=0$ for $z<0$, $F_Z(z)=(1+z)/2$ for $0\le z<1$, and $F_Z(z)=1$ for $z\ge 1$.

For $B_n=\{X_n>a_n\}$, the definition of the common cumulative distribution function givesThe strict exceedance matters here: an atom of a measure at $a_n$ belongs to $\{X_n\le a_n\}$, so it is already included in $F(a_n)$. The events $B_n$ are mutually independent events, since each depends on a separate member of the independent random variables $X_n$.

If ${\sum }_n(1-F(a_n))<\infty$, countable subadditivity gives, for every $N$,Continuity of probability then gives $\mathbb{P}({\mathrm{limsup}}_n{B}_n)=0$. This is the First Borel-Cantelli lemma and does not need independence.

If the series diverges, independence and $1-u\le e^{-u}$ give, for every fixed $N$,Taking $M\to \infty$ shows that the probability of no exceedance after $N$ is zero. A countable union over $N$ shows that finitely many exceedances has probability zero. This is the Second Borel-Cantelli lemma. Together with the preceding event equality, it proves

The conditions on the cumulative distribution function imply $0<X_n\le 2$ almost surely. With ${\mathcal{H}}_n=\sigma (X_1,\dots ,X_n)$, adjoin $Y_0=1$ and the trivial ${\mathcal{H}}_0$. The random variables $Y_n$ are nonnegative, adapted, and bounded by $2^n$ at each fixed time. Independence and $\mathbb{E}{X}_{n+1}=1$ giveThus this is the product martingale from independent mean-one factors, with $\mathbb{E}{Y}_n=1$. The almost sure supermartingale convergence theorem states that a nonnegative supermartingale converges almost surely to a finite integrable random variable. Applied to this nonnegative martingale, it shows $Y_n\to Y_{\infty }<\infty$ almost surely. Every convergent real sequence is bounded, so ${\sup }_{n\ge 1}{Y}_n<\infty$ almost surely.

The printed maximum can in fact be taken literally here. If $X_1=1$ almost surely, all products equal $1$. Otherwise the strict Jensen equality criterion, applied to the strictly convex function $t\mapsto -\sqrt{t}$ on $(0,\infty )$, gives $q=\mathbb{E}\sqrt{X_1}<1$. Independence yields $\mathbb{E}\sqrt{Y_n}=q^n$. By Markov's inequality, ${\sum }_n\mathbb{P}(Y_n>\epsilon )\le {\epsilon }^{-1/2}{\sum }_n{q}^n<\infty$ for every $\epsilon >0$. The First Borel-Cantelli lemma and the countable choices $\epsilon =1/m$ give $Y_n\to 0$ almost surely. Since $Y_1>0$, a sequence with this limit attains its positive supremum among finitely many initial terms. This is an instance of fractional moment decay of independent mean-one products.

Consider the Borel setIts indicator function is nonnegative and measurable, so the Tonelli theorem applies to the product measure, without needing any separate absolute-integrability argument. At a fixed $y\in (a,b]$, its section in the first coordinate is $(a,y]$; outside $(a,b]$ that section is empty. ThereforeThe last equality uses the definition of a cumulative distribution function on the half-closed interval. It retains all point atoms of a measure on the diagonal $x=y$, excludes a possible point atom of a measure at $x=a$, and includes $y=b$, exactly as required by the set $D$. If $a=b$, both sides vanish. This verifies the identity even when the probability measures are not non-atomic measures.^[ref]

Let $F_{n-1}$ be the cumulative distribution function of the Student t-distribution. The two-sided p-value is $2[1-F_{n-1}(|t_0|)]$. If it is less than $\alpha$, strict monotonicity gives $|t_0|>t_{n-1,1-\alpha /2}$, henceThe magnitude of the interval's centre is larger than its half-width, so zero lies outside the confidence interval. This is the duality between confidence intervals and hypothesis tests.