Prove that $X_n/\lambda_n \to$ 1 in probability for $X_n \sim \text{Pois}(\lambda_n)$

Question

Let $X_n \sim \text{Pois}(\lambda_n)$, where $\lambda_n \to \infty$ as $n \to \infty$. Prove that $X_n/\lambda_n \to 1$ in probability

Should I solve this problem using chebychev's inequality? I'm not sure if setting $$P(\lvert X_n-\lambda_n\rvert\geq \epsilon) \leq \frac{\text{Var}(X_n)}{\epsilon^2}=\frac{\sigma^2}{n(\epsilon^2)}$$ would help "prove" this relationship even though the term on the right side goes to zero as $n \to \infty$.

Answer 1

Chebyschev's inequality is indeed a very good idea, but you mixed up several things: First of all, we would like to estimate

$$p:= \mathbb{P} \left( \left| \frac{X_n}{\lambda_n}-1 \right| \geq \epsilon \right)$$

(and not $\mathbb{P}(|X_n-\lambda_n| \geq \epsilon)$.) Obviously,

$$p = \mathbb{P}(|X_n-\lambda_n| \geq \epsilon \lambda_n)$$

and therefore it follows from the Chebyshev inequality that

$$p \leq \frac{1}{(\epsilon \lambda_n)^2} \text{var} \, (X_n) = \frac{1}{\epsilon^2} \frac{1}{\lambda_n}.$$

Letting $n \to \infty$ finishes the proof.