Let lambda>0, and for every n ge 1, we have X_n∼Geo(lambda/n). We define hatX_n=1/n X_n. Show that for every t ge 0: F_{hatX_n}(t) rightarrow F_hatX(t) when n rightarrow infty. while hatX∼Exp(lambda).

Probability question that links between geometric and exponential distribution with limits.
Let $\lambda >0$, and for every $n\ge 1$, we have ${X}_{n}\sim \mathrm{Geo}\left(\frac{\lambda }{n}\right)$.
We define ${\stackrel{^}{X}}_{n}=\frac{1}{n}{X}_{n}$. Show that for every $t\ge 0$:
${F}_{{\stackrel{^}{X}}_{n}}\left(t\right)\to {F}_{\stackrel{^}{X}}\left(t\right)$ when $n\to \mathrm{\infty }.$. while $\stackrel{^}{X}\sim \mathrm{Exp}\left(\lambda \right)$.
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canhaulatlt
Step 1
Answer: $\mathbb{P}\left[\stackrel{^}{{X}_{n}}\le t\right]=\mathbb{P}\left[{X}_{n}\le n\cdot t\right]=\mathbb{P}\left[{X}_{n}\le ⌊n\cdot t⌋\right]=1-\left(1-\frac{\lambda }{n}{\right)}^{⌊n\cdot t⌋}\to 1-{e}^{-\lambda t}={F}_{\stackrel{^}{X}}\left(t\right)$
Step 2
Explanation:
The first equality comes from the definition of $\stackrel{^}{{X}_{n}}$. For the second equality we require the floor function, ⌊⋅⌋, 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 $⌊4.5⌋=4$. This is simply the CDF of a geometric RV with rate $\frac{\lambda }{n}$. For the limit we use: $\underset{n\to \mathrm{\infty }}{lim}\left(1+\frac{x}{n}{\right)}^{n}={e}^{x}$. Recognition of the CDF of Exponential CDF, which I see you have already done.
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Chloe Arnold
Step 1
Let $t>0$ then,
${F}_{\stackrel{^}{{X}_{n}}}\left(t\right)=P\left({X}_{n}/n\le t\right)=P\left({X}_{n}\le tn\right)=\sum _{k=1}^{⌊tn⌋}P\left({X}_{n}=k\right)={p}_{n}\sum _{k=1}^{⌊tn⌋}{q}_{n}^{k-1}$
Where ${p}_{n}=\frac{\lambda }{n}$ and ${q}_{n}=1-{p}_{n}$.
Therefore, ${F}_{\stackrel{^}{{X}_{n}}}\left(t\right)={p}_{n}\frac{1-{q}_{n}^{⌊tn⌋}}{1-{q}_{n}}=1-{q}_{n}^{⌊tn⌋}$
Step 2
Now, ${q}_{n}^{⌊tn⌋}={\left(1-\frac{\lambda }{n}\right)}^{⌊tn⌋}=\mathrm{exp}\left(⌊tn⌋\mathrm{ln}\left(1-\lambda /n\right)\right)$
But as $t>0$ we have,
$⌊tn⌋\sim tn$
and $\mathrm{ln}\left(1-\lambda /n\right)\sim -\lambda /n$
Therefore, $⌊tn⌋\mathrm{ln}\left(1-\lambda /n\right)\sim -\lambda t$
Hence, $⌊tn⌋\mathrm{ln}\left(1-\lambda /n\right)\to -\lambda t$
By continuity of exp we have, ${q}_{n}^{⌊tn⌋}\to {\mathrm{e}}^{-\lambda t}$
Therefore, ${F}_{\stackrel{^}{{X}_{n}}}\left(t\right)\to 1-{\mathrm{e}}^{-\lambda t}={F}_{\stackrel{^}{X}}\left(t\right)$
This is also true for $t=0$, indeed ${F}_{\stackrel{^}{{X}_{n}}}\left(0\right)=0=1-\mathrm{exp}\left(-\lambda 0\right)={F}_{\stackrel{^}{X}}\left(0\right)$