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Substituting the fixed-tension partition function of a one-dimensional chain into the given free-energy definition givesDifferentiation with respect to the parameter conjugate to extension givesThe same result follows directly from the expected value under $p_i=e^{\tau l_i}/Z$:Consequently

Writing $Y_i$ for team1_goal and $x_i$ for team1_xg, the first hypothesis concerns only the expected value:The second specifies the entire conditional probability distribution:with independence between matches as assumed in the question. The Poisson limit theorem, often called the law of small numbers in this setting, suggests a Poisson distribution when a goal count is the sum of many approximately independent rare scoring opportunities.

First suppose $L\in \mathbf{RP}\cap \mathbf{co}\text{-}\mathbf{RP}$. Run the RP algorithm for $L$ and the RP algorithm for its complement with fresh random bits. If the first accepts, output one; if the second accepts, output zero; otherwise repeat. Neither output can be wrong, and on every input the appropriate algorithm accepts in each round with probability at least $1/2$. The number of rounds is dominated by a geometric distribution of mean two, so this is an always-correct algorithm with polynomial expected running time.

Conversely, let $T$ be always correct when it halts and have expected running time at most $p(n)$. Run it for $2p(n)$ steps. Markov inequality givesAccept exactly when $T$ halts and outputs one; this is an RP algorithm for $L$. Accepting exactly when it halts and outputs zero is an RP algorithm for the complement. Thereforeis equivalent to zero-error expected polynomial time.

The minimum of the recovery and death clocks has an exponential distribution with rate $\mu +\gamma$. Its expected value is therefore

Write $Y_i$ for the colony count and $d_i$ for the dose on plate $i$. The first fit is the Poisson regressionThus its log-likelihood isThe Poisson deviance relative to the saturated model iswhere a summand with $y_i=0$ uses $0\log 0=0$.

At dose zero the fitted expected value is $e^{{\hat{\beta }}_0}=e^{3.321995}\simeq 27.72$ eradicated colonies. Increasing dose by one unit multiplies the fitted mean by $e^{{\hat{\beta }}_1}=e^{0.0001901}\simeq 1.000190$; for example, an increase of $100$ units multiplies it by about $1.0192$. The positive fitted effect is small and its displayed two-sided $p$-value, $0.105$, gives little evidence against a zero dose coefficient.

For $u\in C^2(\overline{D})$, Itô formula shows thatis a martingale. Take expectations and let $t\to \infty$. The function $u$ is bounded on the compact set $\overline{D}$, while $\Delta u$ is bounded and ${\mathbb{E}}_{x}T<\infty$ by part (a). The dominated convergence theorem therefore gives Dynkin formula for Brownian motion

Let $X_t=N_t^{+}-N_t^{-}$, where $N^{+}$ and $N^{-}$ are independent Poisson processes of rate $\lambda$. This symmetric Poisson difference process is centered, has jumps $\pm 1$, and has variance rate ${\sigma }^2=2\lambda$. If $a,b$ are positive integers, then $X_T\in \{-a,b\}$ exactly. Optional sampling of the martingale $X_t$ givesConsequentlyPart (d) now gives the mean exit time

For $\mu >0$, let $Q$ be the geometric distributionIf $P$ is the law of $X$, Gibbs inequality givesThereforeFor $\mu =0$, the nonnegative random variable $X$ is zero almost surely and both sides vanish. This proves that the maximum entropy distribution on the nonnegative integers with fixed expected value is geometric.

First the expected value of the absorption time is finite. From any interior state, consecutive leftward moves lead to absorption within at most $n$ steps, with probability at least $p=3^{-n}$. By independence of the subsequent steps, the probability of surviving $jn$ steps is at most $(1-p{)}^j$. Summing this geometric series bounds the expected value of the absorption time by $n/p$.

Now apply first-step analysis. Starting from $k$ in the interior, one step is always used. Conditional on that step, the remaining expected values are $e_{k+1}$, $e_k$, and $e_{k-1}$, each with probability $1/3$. The law of total expectation therefore givesThe term $1$ remains even when the particle stays at $k$, because a step has elapsed. Rearranging gives

Under the extra hypothesis that $G(s)$ exists for every real $s$, its power series has infinite radius of convergence. Termwise differentiation of a power series is therefore valid to every finite order. For $k\ge 1$,At $s=1$ this is the expected value of the falling factorial of $X$:In particular $\boxed{G^{\prime }(1)=\mathbb{E}[X]}$. The terms with $j<k$ vanish because their falling factorials contain a zero factor. These factorial moments are finite under the stated hypothesis.

For a nonnegative integer-valued random variable, its probability generating function isThe convention $s^0=1$, including at $s=0$, gives $G(0)=p(0)$. The power series has absolute convergence for every real $|s|\le 1$, since ${\sum }_{j}p(j)|s{|}^j\le 1$. More generally its real domain as a finite expected value is $\{s\in \mathbb{R}:{\sum }_{j}p(j)|s{|}^j<\infty \}$, which may extend beyond $[-1,1]$. No extension beyond that guaranteed interval is automatic.

Computing the same expected value directly from the Poisson distribution instead givesFor each fixed positive integer $n$, the ratio test shows that this sum converges. This identity is Dobiński's formula. The case $n=0$ also gives $B(0)=1$ if $0^0$ is interpreted as one in this particular moment formula.

The probability density function must integrate to one, soAlso, linearity of expectation and the mean and variance of the standard normal distribution giveUsing the probability density function to compute this expected value givesThese are the values $\Gamma (1/2)$ and $\Gamma (3/2)$ of the gamma function.

By independence, the expected value of the product factors:

Absolute integrability now requiresFor $\alpha \le 0$, the integrand is bounded by $z{e}^{-z^2/2}$, whose integral is finite. For $\alpha >0$, once $z^2\ge 1/\alpha$, the exponent is at least $\alpha z^4/2$, so the integral diverges. HenceAgain the zero follows from the odd function and symmetry; outside the integrable range both signed parts diverge, so the expected value is undefined.

An expected value of a signed random variable is finite precisely when its absolute value is integrable. With the standard normal distribution probability density function,For $\alpha <1/2$, this integral equals $1/[\sqrt{2\pi }(1/2-\alpha )]<\infty$. For $\alpha \ge 1/2$, its integrand does not decay sufficiently and the integral diverges. ThereforeThe value is zero by the odd function and the symmetric distribution. For the other $\alpha$, both positive and negative parts have infinite integrals; formal cancellation would give only a Cauchy principal value, not an expected value. This distinction is the integrability threshold for exponential powers of a Gaussian.

For a geometric distribution on positive integers, the probability generating function is $ps/[1-(1-p)s]$. Conditional averaging over the red die yieldsAs a power series, its exact domain of absolute convergence isIndeed the slowest decaying component has failure probability $5/6$, so $\mathbb{P}(T=t)\ge (1/36)(5/6{)}^{t-1}$. At $|s|=6/5$, the terms of the power series fail to tend to zero; outside that disk they grow. All six component series converge inside it. In particular, the probability generating function is defined on the usual real interval $0\le s\le 1$, with $G_T(1)=1$. The rational expression beyond the convergence disk is only an algebraic continuation, not the defining expected value.

Let $R$ denote the red die. Conditional on $R=r$, each white roll succeeds independently with probability $p_r=(7-r)/6$. Consequently $T\mid R=r$ has the geometric distribution on $\{1,2,\dots \}$ and expected value $1/p_r=6/(7-r)$. The law of total expectation givesThis is $2.45$ rolls; the averaging must be over conditional expected values, not over success probabilities before taking a reciprocal.

Compute fitted responses ${\hat{y}}_i=\hat{\alpha }+\hat{\beta }{x}_i$ and regression residuals $e_i=y_i-{\hat{y}}_i$. A conventional residual plot shows $e_i$ against ${\hat{y}}_i$ or $x_i$; plotting against observation order additionally looks for runs. One checks for curvature, changing spread and other structure rather than random scatter around zero.

An ordinary residual plot need not reveal this failure of independence. The errors have constant mean zero, constant variance one and zero pairwise covariances, so the usual trend and changing-variance clues need not appear. A visually patternless residual plot cannot validate independence; fitted regression residuals themselves also share the fitted coefficients even when the true errors are independent.

There is detectable dependence in the magnitudes. The Gaussian fourth moment is $\mathbb{E}{\eta }_i^4=3$, giving for every adjacent pairFor $i=1$, the product is ${\eta }_1^4{\eta }_2^2$; for $i\ge 2$, it is ${\eta }_{i-1}^2{\eta }_i^4{\eta }_{i+1}^2$, with the same expected value 3. Large magnitudes tend to occur together. A lagged plot of squared regression residuals or a test directed at dependence in their squares is more informative. Thus recognition is possible with suitable diagnostics, but is not guaranteed by the standard residual plot requested here.

Include the negative atom as well as the continuous part in the expected value: