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The adapted process is bounded by $1$, so every $X_n$ is an integrable random variable, indeed a square-integrable random variable. Independence of $U_{n+1}$ from ${\mathcal{F}}_n$ and its continuous uniform distribution giveThis calculation with a random threshold is justified by conditioning on the ${\mathcal{F}}_n$-measurable value of $X_n$: for a fixed $t\in [0,1]$, the Lebesgue integral ${\int }_0^1{1}_{\{u\le t\}}du$ equals $t$. More formally, independence integrates the bounded Borel measurable function $(t,u)\mapsto 1_{\{u\le t\}}$ over its second coordinate. Thereforewhich verifies the martingale property for this filtration.

Since $B_{n+1}^2=B_{n+1}$, the conditional expectation of the squared martingale difference isThe law of total expectation yields $\mathbb{E}[(X_{n+1}-X_n{)}^2]=\frac{1}{4}\mathbb{E}[X_n(1-X_n)]$. In particular, $\mathbb{E}{X}_n=x$ for every $n$.

For each $n$,since $S_k$ is adapted. This verifies directly that $\tau$ is a stopping time.

Set $d=b-a$. On $\{\tau >md\}$, the random walk is at some integer strictly between $a$ and $b$. If its next $d$ steps are all $+1$, it must hit $b$ within that block. These future steps are independent of ${\mathcal{F}}_{md}$ and have probability $2^{-d}$ of being all $+1$. HenceThe law of total expectation and mathematical induction give the geometric tail bound from a uniform escape probability:Therefore $\mathbb{P}(\tau <\infty )=1$. In fact this also proves $\mathbb{E}\tau \le d{2}^d<\infty$ by grouping the tail-sum formula into blocks of length $d$. No assertion of independence between successive survival events is needed.

The function ${\varphi }_a(t)=t\wedge a$ is a concave function, and $X\wedge a$ is integrable since $|X\wedge a|\le |X|+a$. The concave form of the conditional Jensen inequality givesThe random variables $X$ and $Y$ have the same probability distribution, so $\mathbb{E}[X\wedge a]=\mathbb{E}[Y\wedge a]$. Using the law of total expectation, the nonnegative difference between the right and left sides of the inequality has zero expectation. It must vanish almost surely. ThereforeThe equality comes from the distributional assumption as well as the conditional Jensen inequality.

Put $S=X+Y$. Given $X$, its conditional expected value is $3X/2$ and its conditional variance is $X/4$, since $X$ is then fixed. The law of total expectation and law of total variance giveThe two scores are dependent: the law of total covariance gives $\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,X/2)=q/(2p^2)$. Adding their two variances without that covariance term would give an incorrect total.

The law of total expectation gives $\mathbb{E}Y=\mathbb{E}[X/2]=q/(2p)$. The law of total variance separates the two sources of randomness:Substitution givesThe binomial thinning of a geometric random variable identity supplies a further check: $Y$ itself has a zero-based geometric distribution with success probability $2p/(1+p)$.

For an integrable random variable $X$ and a partition of a sample space $(B_i)$, the law of total expectation statesIt follows by writing $X={\sum }_{i}X{1}_{B_i}$ and using $\mathbb{E}[X{1}_{B_i}]=\mathbb{P}(B_i)\mathbb{E}[X\mid B_i]$. Integrability justifies the countable sum. The same equality is valid with nonnegative, possibly infinite expectations when $X\ge 0$.

Let $\tau$ be the absorption time. Its expectation is finite: from every interior state, a string of at most $n$ successive left moves reaches $0$, and the probability of such a string is at least $p=3^{-n}$. Hence the Markov property gives ${\mathbb{P}}_k(\tau >jn)\le (1-p{)}^j$, a geometric tail bound, and ${\mathbb{E}}_k[\tau ]<\infty$.

For $1\le k\le n-1$, one step is always spent before absorption or the remaining walk. Apply first-step analysis and the law of total expectation to the three possible first moves. The conditional expectations of the remaining absorption time are $e_{k+1},e_k,e_{k-1}$, respectively; independence of future moves gives the same law as a fresh random walk from the new position. Thuswhich rearranges toThe term $1$ also appears when the walk stays at $k$, because staying still consumes a step.

For $0\le s\le 1$, condition on $N=n$. Its independence from the $X_i$ preserves their independent common distribution, so part (ii) givesFor $n=0$, this is $1$, agreeing with $Y=0$. By the law of total expectation,This proves the probability generating function of an independent random sum. For nonnegative $s$ outside $[0,1]$, the same argument is valid whenever the right side is finite, by Tonelli's theorem for nonnegative sums. For signed or complex $s$, a sufficient absolute-convergence condition is $G_N(G_X(|s|))<\infty$; it holds automatically on $|s|\le 1$. Outside this domain, mere evaluation of an analytically extended expression is not a substitute for existence of the probability generating function.

Given $M=m$, there are exactly $m$ independent opportunities to stay awake, each with probability $p$. HenceThe standard binomial distribution formulas are $\mathbb{E}[W\mid M]=pM$ and $\mathrm{var}(W\mid M)=Mp(1-p)$. The Poisson distribution formulas are $\mathbb{E}[M]=\lambda$, $\mathrm{var}(M)=\lambda$, and therefore $\mathbb{E}[M^2]={\lambda }^2+\lambda$. By the law of total expectation,Using the definition of covariance,Thus $W$ and $M$ are not independent, even though the separate awake and asleep counts produced by Poisson thinning are independent.

For an integrable random variable $X$, the law of total expectation on the finite or countable measurable partition isHere $\mathbb{E}(X\mid B_i)=\mathbb{E}[X{1}_{B_i}]/\mathbb{P}(B_i)$ for positive-probability cells; null cells contribute zero and need no conditional value. The series is absolutely convergent since its absolute terms sum to at most $\mathbb{E}|X|$. For a nonnegative $X$, the same identity is valid with possible value $+\infty$.

Use the measurable partition $B_n=\{N=n\}$ for $n\ge 0$, omitting any null cells. For $0\le s\le 1$, the random variable $s^X$ is bounded, so the law of total expectation applies without moment assumptions. Conditional on $N=n$, the sum contains exactly $n$ terms. Independence of $N$ from the whole sequence preserves their joint law under this conditioning; independence of the $Y_k$ then givesFor $n=0$ the empty sum is zero and the empty product is one, in agreement with the formula. Therefore the probability generating function of an independent random sum is

The conditional expectation given an event of positive probability isApply the law of total probability to each event $\{X=j\}$:This is the law of total expectation. Interchanging the sums is justified by nonnegative terms, for example by Tonelli theorem, so it remains valid when the expectation is infinite.

Let $J$ count successful journeys before the first breakdown. The probability of surviving the first $r$ journeys is the tail probability $\mathbb{P}(J\ge r)$. For $d\ge 0$, independence giveswhere again $q=(1-\alpha )(1-\beta )$. For a nonnegative integer-valued random variable, the tail-sum formula follows by writing $J={\sum }_{r\ge 1}{1}_{\{J\ge r\}}$ and interchanging nonnegative sums. ThereforeAs a separate first-day check using the law of total expectation, if $E=\mathbb{E}(J)$, then $E=\alpha \cdot 0+(1-\alpha )\beta \cdot 1+q(2+E)$. Solving gives the same finite value. This counts only successful journeys, excluding the failed journey.

Condition first on $X$. A Poisson distribution with parameter $k$ has conditional probability generating function $e^{k(s-1)}$, including the degenerate zero case when $k=0$. The law of total expectation givesFor $0\le s\le 1$ the inner argument is in $(0,1]$, so this use of the generating function is justified directly. The result is a compound Poisson distribution with a Poisson distribution of unit parameter for each jump, also called a Neyman Type A distribution in this case. It is generally not an ordinary Poisson distribution.

The conditional mean of a Poisson distribution equals its parameter, so $\mathbb{E}[Z\mid X]=X$. The law of total expectation givesAs a check, differentiating the probability generating function in part (iii) at one also gives $G_Z^{\prime }(1)=\mu$.

The finite-partition form of the law of total expectation statesBecause the events form a partition of a sample space of the sample space, ${\sum }_j{1}_{B_j}=1$ pointwise. Multiply by $X$ and use linearity of expectation to obtainAn event of zero probability contributes zero to the middle expression and is omitted from the final expression, so no undefined conditional expectation is multiplied by zero.

Complete the square in the joint probability density function:The density therefore factors as the standard normal distribution density for $Y$ times a normal distribution density in $x$ with mean $\rho y$ and variance $1-{\rho }^2$. It follows that $\mathbb{E}[Y]=0$, $\mathbb{E}[Y^2]=1$, and $\mathbb{E}[X\mid Y]=\rho Y$. Symmetry gives $\mathbb{E}[X]=0$. By the law of total expectation,Consequently the covariance of this bivariate normal distribution is $\mathrm{Cov}(X,Y)=\rho$. This calculation derives the result from the given density.

A partition of a sample space is a finite or countable family of pairwise disjoint events $B_i$ whose union is the whole sample space. For an integrable random variable, the partition form of the law of total expectation isZero-probability cells contribute zero and are omitted because elementary event-conditioning on them is undefined. The formula also holds with extended nonnegative values when $X\ge 0$. Integrability makes the signed series legitimate in the finite-expectation case.

Apply the law of total expectation to the bounded random variable $R_{k+1}$. Linearity of expectation givesThis also holds for the first step from $r_0=n$.

Choose two hypothetical environments A and B of one chemical element, with line centers ${\mu }_A=284\mathrm{eV}$ and ${\mu }_B=286\mathrm{eV}$. Model each line by a normalized Gaussian function with common standard deviation $\sigma$, and stipulate integrated count weights $w_A=3/4$, $w_B=1/4$. On the binding energy axis $E$, the normalized count density isAll energies in this expression use electronvolts. Each component's integrated area equals its count weight, and $\int I(E)dE=1$. There is no background, statistical counting fluctuation, or additional emission feature in this stipulated model. These choices deliberately isolate photoelectron peak overlap and resolution.

Direct integration, or the law of total expectation applied to the component label, gives the centroidThe centroid lies between the chemical environments and is not either component's binding energy. Expanding $(E-\bar{E}{)}^2$ within each component, the odd cross term integrates to zero. The resulting variance isFor $\sigma =0.25\mathrm{eV}$ the variance is $0.8125{\mathrm{eV}}^2$ and the two maxima are separated. For $\sigma =1.00\mathrm{eV}$ it is $1.7500{\mathrm{eV}}^2$ and there is one maximum. Each Gaussian's full width at half maximum is $2\sqrt{2\ln 2}\sigma$, approximately $0.589$ and $2.355\mathrm{eV}$ respectively. The chemical separation remains $2\mathrm{eV}$ in both cases. The number of visible maxima therefore does not equal the number of chemical components.

One interpretation of this change is additional independent Gaussian instrument blur. The convolution of probability densities adds the variances, so broadening $\sigma =0.25$ to $1$ requires response standard deviation $\sqrt{1-0.25^2}=\sqrt{15}/4\simeq 0.968246\mathrm{eV}$. A normalized response with zero mean preserves the centroid and integrated counts. Real line shapes and backgrounds can depart from this model, and separating hidden components is an inverse problem; a plausible fit does not alone establish uniqueness.

Finally, count weights describe detection. In a simple response model, let $n_i$ denote the number of emitting sites sampled in environment $i$, $s_i$ its effective photoionization cross section, and $T_i$ the probability that its emitted electron contributes to the selected measured line. ThenThe common factor $C$ includes exposure. Here $T_i$ includes escape from the sample, analyzer transmission and detection. Count weights equal site fractions only if the relevant response factors are equal or corrected. They are not automatically molecular mole fractions; different molecules can supply different numbers of the selected emitting sites. In an original example, equal site numbers with $s_A{T}_A=3s_B{T}_B$ give the plotted $3:1$ count ratio despite a $1:1$ site ratio.