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For independent exponential distributions with rates ${\lambda }_1>{\lambda }_2>0$, let $T=S_1+S_2$. Dividing the joint density of $(S_1,T)$ by the hypoexponential distribution density of $T$ gives a truncated exponential distribution on $(0,t)$ with rate ${\lambda }_1-{\lambda }_2$. Its displayed conditional expectation supplies an E-step when only pair totals are observed. At equal rates the conditional split is uniform, with mean $t/2$.

// Target: probability-and-statistics.bigb

At saturating ATP, binding need no longer dominate the observed dwell, but saturation alone does not determine a dwell-time distribution. The standard interpretation of a distribution that rises from zero is that at least two sequential events remain rate limiting. Model a dwell as $T=T_1+T_2$, where the independent waiting times are distributed according to exponential distributions with rates $k_1,k_2$. Their convolution isThis hypoexponential distribution starts at zero, rises to a mode $\ln (k_2/k_1)/(k_2-k_1)$, then has an exponential tail. When the rates are equal, the continuous limit is the two-stage Erlang distribution,A single slow memoryless event instead gives $k{e}^{-kt}$, maximal at zero. Three 120-degree steps per revolution specify rotational stoichiometry, not three slow reactions within every dwell. The two-event interpretation is supported by the resolved catalytic-dwell chemical kinetics in the primary F1 experiment.^[ref]

The fit assumes sequential approximately irreversible events, constant rates during each dwell, negligible ATP-binding delay, negligible mechanical transit time, and detection of all relevant steps. Parallel pathways, backward transitions, inhibited states, probe drag and missed short dwells can change the histogram. The mathematical fit alone does not identify each event chemically or prove that two ATP molecules are hydrolyzed per step.

The given probability density function is the completion-time density for two independent sequential irreversible steps, with exponential distributions of waiting times at rates $k_a,k_b>0$. Let $T=T_a+T_b$. The convolution isIt is a two-stage hypoexponential distribution. The density vanishes at zero, initially rises as $k_a{k}_{b}t$, peaks, and then decays with a tail governed by the slower rate. Setting its derivative to zero givesNormalization follows either from the convolution of normalized densities or directly from $(k_a{k}_b/(k_a-k_b))(1/k_b-1/k_a)=1$. In the equal-rate limit $k_a=k_b=k$, the apparent singularity is removable andThis is an Erlang distribution with shape two. If one rate is much faster, its short waiting time becomes hard to resolve and the distribution resembles a single exponential distribution after its initial rise.

The observation supports at least two kinetically resolved stages before the recorded event, rather than one memoryless rate-limiting event. Possibilities include substrate binding followed by a conformational or catalytic step. It does not identify which stage is binding, chemistry, product release or rotation. The density is unchanged by exchanging $k_a$ and $k_b$, so temporal order and chemical identity cannot be recovered from this fit alone. Other hidden-state mechanisms can also mimic the same density. Dependence of one fitted rate on ATP concentration, for example, would provide additional evidence about assignment, rather than following from the distribution by itself.

The left panel illustrates b for an initially closed population. The other panels are dimensionless examples of the distributions derived here; none reconstructs the missing source trace.

The sum of the two independent exponential times has a hypoexponential distribution. By convolution of probability densities, for $t>0$ and $d={\lambda }_1-{\lambda }_2>0$,Thereforeand $g(t)=0$ for $t<0$. This is a nonnegative normalized probability density function. At the equal-rate boundary it tends to ${\lambda }^{2}t{e}^{-\lambda t}$, a shape-two Gamma distribution.

Put $A={\sum }_{i=1}^3{S}_{i1}$, $B={\sum }_{i=1}^3{S}_{i2}$ and $d={\lambda }_1-{\lambda }_2$. Independence of the pairs makes their likelihood contributions multiply. The three fully observed pairs contribute $({\lambda }_1{\lambda }_2{)}^3{e}^{-{\lambda }_{1}A-{\lambda }_{2}B}$; each remaining pair contributes its hypoexponential distribution density. Thus the observed-data log-likelihood isAn equivalent expression, convenient for calculations, isThe apparent singularity as $d\downarrow 0$ cancels between the last sum and $-5\log d$. The resulting equal-rate limit is finite for positive observed times.

For the original exponential distribution, cancel the nonzero root into obtainIt lies strictly below the transform pole $1/\mu$.

With the extra expenses, the insurer's payment per claim is $Y=X+A$, where $A$ has exponential distribution of expected value $\mu /2$ and is independent of $X$. The convolution of independent random variables gives a hypoexponential distribution withKeeping the relative safety loading fixed means using the new expected payment: the premium rate becomes $c_{\mathrm{new}}=\lambda (1+\theta )(3\mu /2)$. It does not mean keeping the old premium rate fixed. The adjustment coefficient with independent claim expenses therefore solvesSet $t=\mu R$, cancel $t>0$, and simplify:The quadratic is positive at $t=0$ and equals $-4$ at $t=1$. Its leading coefficient is positive, so the smaller root is in $(0,1)$ and the larger root exceeds $1$. Only the smaller root lies in the finite moment-generating function domain. HenceFor $\theta =1$,Thus the new adjustment coefficient is about $9.72\%$ smaller, even though the premium rate has been increased to retain the same relative safety loading. The Lundberg inequality consequently has a slower exponential decay rate as a function of capital.

Use the rate convention for the exponential distribution. By convolution of independent random variables, the probability density function of $S=X+Y$ is, for $s\ge 0$,It is zero for $s<0$. The expression is nonnegative whichever rate is larger, and defines a two-stage hypoexponential distribution. As a check, its limit as $\mu \to \lambda$ is ${\lambda }^{2}s{e}^{-\lambda s}$, the shape-two gamma distribution.

For $M=\min \{X,Y\}$, independence gives the survival function $\mathbb{P}(M>s)=e^{-\lambda s}{e}^{-\mu s}$ for $s\ge 0$. Differentiating its complement givesThus the minimum has an exponential distribution whose rate is the sum of the rates.

Two sequential independent memoryless reaction waits have a hypoexponential distribution. Convolution gives $p(t)=k_1{k}_2(e^{-k_{1}t}-e^{-k_{2}t})/(k_2-k_1)$. Equal rates give the Erlang distribution $k^{2}t{e}^{-kt}$. A nonzero mode distinguishes this ideal two-event model from a single exponential wait, but does not chemically identify the events.