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The surplus in the classical risk model is $U(t)=u+ct-S(t)$. With ruin defined as a strictly negative surplus, its ultimate ruin probability isLet ${\overline{F}}_X(v)={\int }_v^{\infty }{f}_X(x)dx$. The first arrival time has exponential distribution of rate $\lambda$, independent of the first claim. Before that claim no ruin occurs, because the premium rate is positive. Given its arrival at $t$ and size $x$, available capital just before it is $u+ct$. If $x>u+ct$ ruin is immediate; otherwise the independent increments and fresh claim sizes make subsequent ruin probability $\psi (u+ct-x)$. Thus first-claim conditioning givesThis is the ruin counterpart of the first-claim decomposition for survival probability. DefineThe convolution term is continuous: extend the bounded function $\psi$ by zero to negative arguments and use continuity of translations in $L^1$ for the integrable density $f_X$. The tail function is also continuous. Changing variables to $v=u+ct$ yieldsDifferentiating this exponentially weighted integral establishes differentiability rather than presuming it. The fundamental theorem of calculus givesReversing the integration variable in the convolution therefore proves the ruin integro-differential equation:At zero the derivative is the right derivative; the same representation supplies it.

For the exponential claim law, introduceDifferentiating the explicit convolution gives $A^{\prime }=\alpha \psi -\alpha A$. Henceusing the proved integral equation. Differentiating that equation once more givesso the required ordinary differential equation isPositive relative safety loading means $c>\lambda /\alpha$, so $\kappa =\alpha -\lambda /c>0$. The general solution is $a+b{e}^{-\kappa u}$.

The condition at large capital must also be used. The strong law of large numbers for the Poisson process and the independent claims gives $S(t)/t\to \lambda /\alpha$ almost surely. Thus $S(t)-ct\to -\infty$. Its path has only finitely many jumps on each bounded time interval, so its supremum over all time is finite almost surely. Therefore $\psi (u)\to 0$ as $u\to \infty$, forcing $a=0$. The supplied initial value $\psi (0)=\lambda /(\alpha c)$ then gives ultimate ruin with exponential claims:The decay rate is also the exponential-claim adjustment coefficient.

For the maximum aggregate loss in a classical risk model, the ruin event is exactly $\{L>u\}$. Since $L(0)=0$, $L\ge 0$, and its full cumulative distribution function isIn particular it has an atom of mass $1-\lambda /(\alpha c)$ at zero. Conditional on $L>0$, its distribution is exponential with rate $\kappa$; omitting that atom would give an incorrectly normalized law.

In the classical risk model, let $N(t)$ be the Poisson process of claim arrivals, independent of the independent claim sizes. The reserve starts at $u$ and earns premiums continuously at rate $c$, soHere $\psi$ is the ultimate ruin probability and $\varphi =1-\psi$ is the survival probability. The relative safety loading is $\theta =(c-\lambda \mu )/(\lambda \mu )>0$.

To derive the survival integro-differential equation, condition on the first claim time $T$, which has the exponential distribution of rate $\lambda$. Immediately before that claim the available capital is $u+cT$. A claim $x\le u+cT$ leaves survival probability $\varphi (u+cT-x)$, since the future claim process restarts with the same independent law; a larger claim causes ruin immediately. The first-claim decomposition for survival probability is thereforePut $r=\lambda /c$ and $H(v)={\int }_0^v\varphi (v-x)f(x)dx$. Changing variables to $v=u+ct$ yieldsThis representation justifies differentiation. The function $H$ is continuous: extend $\varphi$ by zero to negative arguments, then use continuity of translations of the integrable density $f$ in $L^1$ to control its convolution with the bounded function $\varphi$. Differentiating the lower endpoint of the displayed integral gives ${\varphi }^{\prime }=r\varphi -rH$. Since $c=(1+\theta )\lambda \mu$,This deduction uses the first-arrival law rather than taking the desired differential equation as an assumption.

For $f(x)=4x{e}^{-2x}$, the claim law is an Erlang distribution of shape two and rate two. Its mean is $\mu =4{\int }_0^{\infty }{x}^2{e}^{-2x}dx=1$. With $\theta =2$, we have $r=1/3$. Changing the convolution variable to $t=u-x$ givesHence the first requested specialization isLet $J(u)={\int }_0^u\varphi (t)e^{2t}dt$. Differentiation under the finite integral gives $I^{\prime }=J$ and $J^{\prime }=\varphi e^{2u}$. Differentiating the preceding equation and using $4e^{-2u}I=\varphi -3{\varphi }^{\prime }$ then givesThus the second identity isDifferentiate once more:The preceding second-derivative equation gives $8e^{-2u}J=-6{\varphi }^{\prime \prime }-10{\varphi }^{\prime }+4\varphi$. Substitution cancels the undifferentiated terms and provesThis is the Erlang claim-size differential equation for survival probability with $\beta =2$ and $r=1/3$.

Its characteristic polynomial factors as $z(3z+8)(z+1)$. The limit $\varphi (u)\to 1$ therefore makes the general bounded solutionThe supplied zero-capital survival probability is $\varphi (0)=2/3$, so $A+B=-1/3$. We also need the initial derivative carried by the original integral equation: $I(0)=0$ implies ${\varphi }^{\prime }(0)=\varphi (0)/3=2/9$, giving $-A-(8/3)B=2/9$. Solving gives $A=-2/5$ and $B=1/15$. Therefore the survival and ruin probabilities areThe extra initial derivative is essential: $\varphi (0)$ and the limit at infinity alone would leave one free parameter in the third-order equation. The result has $\psi (0)=1/3$ and tends to zero. It is positive and decreasing, since ${\psi }^{\prime }(u)=-(2/5)e^{-u}+(8/45)e^{-8u/3}<0$ for $u\ge 0$. These signs agree with its interpretation as an ultimate ruin probability.

Use the classical risk model surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with $c>0$, and write $\phi$ for its ultimate survival probability. The first-claim decomposition for survival probability follows by conditioning on the first arrival time of the Poisson process. That time has exponential distribution of rate $\lambda$. A claim at time $t$ leaves capital $u+ct-x$ if $x\le u+ct$, after which the Markov property restarts the same risk model. ConsequentlyPut $C(v)={\int }_0^v\phi (v-x)f(x)dx$ and change variable $v=u+ct$:The convolution $C$ is continuous since $\phi$ is bounded and $f$ is integrable. Differentiating proves the survival integro-differential equation

For the specified claim law, expand its probability density function as $f(x)=\frac{1}{2}{e}^{-x}+\frac{1}{4}{e}^{-x/2}$. This is an equal-weight mixture distribution of exponential distributions with rates $1$ and $1/2$. In the convolution, set $z=u-x$ to obtainSince $\lambda /c=10/21$, the requested coefficients are

For the exponential-mixture differential equation for survival probability, define $A(u)=e^{-u}{I}_1(u)$, $B(u)=e^{-u/2}{I}_2(u)$ and $r=\lambda /c$. Differentiating the integrals yieldsApply $(D+1)(D+\frac{1}{2})$ to the last equation, where $D=d/du$. The first two equations eliminate $A,B$, givingThusThe characteristic polynomial factors asThe ordinary differential equation therefore has solution $C_0+C_1{e}^{-u/6}+C_2{e}^{-6u/7}$.

The claim expected value is $\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 2=3/2$. The relative safety loading is $\rho =2.1/(3/2)-1=2/5$, and the given zero-capital survival probability is $\phi (0)=2/7$. The limit at infinity sets $C_0=1$. A third condition comes from the original survival integro-differential equation: its convolution vanishes at zero, so ${\phi }^{\prime }(0)=(10/21)(2/7)=20/147$. Thereforewhich gives $C_1=-20/29$ and $C_2=-5/203$. The final survival probability and ultimate ruin probability areBoth exponential coefficients of $\psi$ are positive. Hence $\phi$ increases from $2/7$ to one, and $\psi$ decreases from $5/7$ to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.

In the classical risk model with premium rate $c>0$ and claim Poisson process rate $\lambda$, the first-claim decomposition for survival probability implies the displayed equation for ultimate survival probability. Change the first-claim integral to an integral over available capital and differentiate its lower limit. The bounded survival probability function convolved with the integrable claim probability density function is continuous, so this also establishes the needed differentiability. In particular ${\phi }^{\prime }(0)=(\lambda /c)\phi (0)$.