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For Borel sets $A_1,\dots ,A_n$, multiply the finite-time Radon-Nikodym derivative into the product law of the increments:The factorization, extended from rectangles to the product sigma-algebra, proves independence and identical distributions. Their common law is the exponential tiltingFor $\lambda >0$, $|x|e^{\lambda x}$ is bounded on $x\le 0$, while $x{e}^{\lambda x}\le C_{\epsilon }{e}^{(\lambda +\epsilon )x}$ on $x\ge 0$. Choosing a small two-sided neighborhood of $\lambda$ still contained in $(0,\infty )$ gives an integrable dominating function for the derivative. Differentiation under the integral sign then yieldsThe mean under one-sided exponential tilting needs an endpoint qualification. At $\lambda =0$, finiteness of positive exponential moments makes $X_1^{+}$ integrable, but gives no integrability of $X_1^{-}$. The formula holds with the right derivative and the extended mean ${\varphi }_{+}^{\prime }(0)=\mathbb{E}{X}_1\in [-\infty ,\infty )$. To see this, apply monotone convergence to $(1-e^{-h{X}_1^{-}})/h$ and dominated convergence to $(e^{h{X}_1^{+}}-1)/h$ as $h\downarrow 0$. A finite mean at zero is not guaranteed by the printed hypotheses: $\mu (\{-j\})=6/({\pi }^2{j}^2)$, $j\ge 1$, has finite $\varphi (\lambda )$ for every $\lambda \ge 0$ but $\mathbb{E}{X}_1=-\infty$.

First take $k$ to be a nonnegative integer. The exponential tilting constructed from the independent increments restricts to ${\mathbb{P}}_n^{\lambda }$ on each ${\mathcal{F}}_n$. Since its Radon-Nikodym derivative is strictly positive, finite-time change of measure givesUnder the tilted law, $L_n=(M_n^{\lambda }{)}^{-1}$ is a martingale, becauseUse the stopped likelihood ratio under exponential tilting at this finite horizon. In detail, $\{{\tau }_k=j\}\in {\mathcal{F}}_j$, soHere the last expression is defined only on the displayed event, so no value of $M_{\infty }^{\lambda }$ is being assumed. The upward skip-free random walk has integer increments bounded above by one, which imply $S_{{\tau }_k}=k$ on a finite hit: the preceding value is at most $k-1$, and there is no upward overshoot. ConsequentlyAt $\lambda ={\lambda }_0$, the mean under one-sided exponential tilting is finite and positive. The strong law of large numbers gives $S_n/n\to {\varphi }^{\prime }({\lambda }_0)>0$ under ${\mathbb{P}}^{{\lambda }_0}$, so every integer level is hit almost surely. Since $\varphi ({\lambda }_0)=1$, passing to $n\to \infty$ givesLet $H={\sup }_{n\ge 0}{S}_n$. The event $\{H\ge k\}$ is exactly the finite-hit event: an integer sequence with all its values below $k$ has supremum at most $k-1$. Also $\mathbb{P}(H=\infty )={\lim }_k{e}^{-{\lambda }_{0}k}=0$. Subtraction of consecutive tail probabilities gives the geometric maximum of an upward skip-free random walk:This is the geometric distribution counting failures before the first success, with success parameter $1-e^{-{\lambda }_0}$.

The PDF writes only $k\ge 0$, without explicitly declaring integer levels. If it is read as allowing real $k$, then ${\tau }_k={\tau }_{\lceil k\rceil }$ and the displayed factor must be $e^{-\lambda \lceil k\rceil }$; the hitting probability is likewise $e^{-{\lambda }_0\lceil k\rceil }$. For example, increments $+1$ and $-1$ with probabilities $1/3$ and $2/3$ have ${\lambda }_0=\log 2$: hitting level $1/2$ has probability $1/2$, not $2^{-1/2}$. The integer interpretation supplies the intended formula.