Suppose that I have two Poisson random variables $X_1$ and $X_2$ with means $\lambda_1$ and $\lambda_2$ respectively. If I know $\lambda_2 > \lambda_1$ then does it follow that $ X_1 \preceq_{st} X_2$?
So far what I have is the following: \begin{align*} X_1 \preceq_{st} X_2 &\iff P \lbrace X_1 > t \rbrace \leq P \lbrace X_2 > t \rbrace ~~~~~~~ -\infty < t < \infty \\ &\iff 1 - e^{-\lambda_1} \sum_{i = 0}^t \dfrac{\lambda^i_1}{i!} \leq 1 - e^{-\lambda_2} \sum_{i = 0}^t \dfrac{\lambda^i_2}{i!} \\ &\iff - e^{-\lambda_1} \sum_{i = 0}^t \dfrac{\lambda^i_1}{i!} \leq - e^{-\lambda_2} \sum_{i = 0}^t \dfrac{\lambda^i_2}{i!} \\ &\iff e^{-\lambda_2} \sum_{i = 0}^t \dfrac{\lambda^i_2}{i!} \leq e^{-\lambda_1} \sum_{i = 0}^t \dfrac{\lambda^i_1}{i!} \\ &\iff \color{red}{\lambda_1 < \lambda_2} ~~~ (*)\\ \end{align*} However I am not sure how solid the ending is. Any help would be greatly appreciated!
$\lambda_1\le\lambda_2 \implies X_1\preceq_{st}X_2$: Suppose that randon variables $X_1, Z$ are independent, $\mathsf{P}(Z\ge0)=1$ and $X_2=X_1+Z$, then $X_1\preceq_{st} X_2$. To prove this just use convolution formula as following $$ F_{X_2}(t)=\int_{-\infty}^{\infty}F_Z(t-s)dF_{X_1}(s)=\int_{-\infty}^{t}F_Z(t-s)dF_{X_1}(s)\le \int_{-\infty}^{t}dF_{X_1}(s)=F_{X_1}(t), \qquad \forall t.$$ Now let $X_1\sim P(\lambda_1), Z\sim P(\lambda_2-\lambda_1)$ where $X_1$ and $Z$ are independent. Then $X_2\sim P(\lambda_2)$ and $$ X_1\preceq_{st} X_2. $$
$X_1\preceq_{st}X_2 \implies \lambda_1\le \lambda_2$: $$ \lambda_1=\mathsf{E}[X_1]=\int_0^\infty F(X_1>t)dt\le \int_0^\infty F(X_2>t)dt=\mathsf{E}[X_2]=\lambda_2.$$