Probability question that links between geometric and exponential distribution with limits.

Question

Let $\lambda>0$, and for every $n\ge 1$, we have $X_n\sim \operatorname {Geo}(\frac{\lambda}{n})$.
We define $\hat X_n=\frac{1}{n}X_n$. Show that for every $t\ge 0$:
$F_{\hat X_n}(t)\to F_{\hat X}(t)$ when $n\to\infty.$ while $\hat X\sim\operatorname {Exp}(\lambda)$.

I know that since $\hat X$ is distributed exponentially: $F_{\hat X}(t)=1-e^{-\lambda t}$.

Trying to understand $\hat X_n$:
$P(\hat X_n = k)=(1-\frac{\lambda}{n})^{k-1}\frac{\lambda}{n}$.
I can see where the $e^{-\lambda}$ will come from when we take the limit, but I'm having some difficulties getting to $F_{\hat X_n}(t)=P(\hat X_n \le t)$, when I find that I can multiply it by $\frac{1}{n}$, and it will be some limits work to reach the proof in my opinion.

I would really appreciate any feedback and help in building my proof.
Thanks in advance!

Answer 1

Answer:

$\mathbb{P}[\hat{X_n} \leq t] = \mathbb{P}[X_n \leq n\cdot t] =\mathbb{P}[X_n \leq \lfloor n\cdot t \rfloor ] = 1-(1-\frac{\lambda}{n})^{\lfloor n\cdot t \rfloor} \to 1-e^{-\lambda t}=F_{\hat X}(t) $

Explaining:

1. The first equality comes from the definition of $\hat{X_n}$
2. For the second equality we require the floor function, $\lfloor \cdot \rfloor$, this is because, as I mentioned in my comment the Geometric distribution takes only integer values. What is the probability of rolling less than or equal to $4.5$ on a dice? The same as rolling less or equal to $\lfloor 4.5 \rfloor =4 $
3. This is simply the CDF of a geometric RV with rate $\frac{\lambda}{n}$
4. For the limit we use: $\lim_{n\to \infty}(1+\frac{x}{n})^n = e^x$
5. Recognition of the CDF of Exponential CDF, which I see you have already done.

If you need any more explaining please comment :)

Answer 2

Thank you for explaining to me the notations.

Let $t >0 $ then,

$$F_{\hat{X_n}}(t) = P(X_n/n\leq t) = P(X_n \leq tn) = \sum_{k=1}^{\lfloor{tn}\rfloor}P(X_n=k)=p_n\sum_{k=1}^{\lfloor{tn}\rfloor}q_n^{k-1}$$

Where $p_n = \dfrac{\lambda}{n}$ and $q_n = 1-p_n$.

Therefore,

$$F_{\hat{X_n}}(t)=p_n \dfrac{1-q_n^{\lfloor{tn}\rfloor}}{1-q_n} = 1-q_n^{\lfloor{tn}\rfloor} $$

Now,

$$q_n^{\lfloor tn \rfloor}=\left(1-\dfrac{\lambda}{n}\right)^{\lfloor tn \rfloor}=\exp(\lfloor tn \rfloor\ln(1-\lambda/n))$$

But as $t>0$ we have,

$$\lfloor tn \rfloor \sim tn$$

and $$\ln(1-\lambda/n) \sim -\lambda/n$$

Therefore,

$$\lfloor tn \rfloor\ln(1-\lambda/n)\sim-\lambda t $$

Hence,

$$\lfloor tn \rfloor\ln(1-\lambda/n) \rightarrow -\lambda t$$

By continuity of $\exp$ we have,

$$q_n^{\lfloor tn \rfloor} \rightarrow \mathrm{e}^{-\lambda t}$$

Therefore,

$$ F_{\hat{X_n}}(t) \rightarrow 1-\mathrm{e}^{-\lambda t} = F_{\hat{X}}(t)$$

This is also true for $t=0$, indeed $F_{\hat{X_n}}(0) = 0 = 1-\exp(-\lambda 0) = F_{\hat{X}}(0)$