Consider the random variable $Y_p=(1-p)X_p$, $P[X_p=k]=(1-p)p^k$. Prove the convergence of the distr. func. of $Y_p$ for $p\to 1$. Consider the random variable $Y_p: \mathbb{N}_0 \to \mathbb{R}$ defined by:
\begin{equation*}
Y_p=(1-p)X_p
\end{equation*}
Here $X_p$ is also a random variable that is geometrically distributed i.e. $P[X_p=k]=(1-p)p^k,\ k\geq 0$. Prove that the distribution function of $Y_p$ converges pointwise to the distribution function of the exponential distribution i.e.
\begin{align*}
F(x):= 1- e^{-x},\ x\geq 0
\end{align*}
for the case $p\to 1$.
This is the task. However, I don't think that the statement in the task is correct (or I am wrong), because: First of all we know that:
\begin{align*}
P[Y_p=k]=(1-p)^2 p^k
\end{align*}
And, thus, the distribution function of $Y_p$ is equal to
\begin{align*}
\mathbb{P}(x)=\sum_{k=0}^\infty (1-p)^2 p^k 1_k(x)
\end{align*}
Thus, if we look at any single point say $x=j$, we get that $\mathbb{P}(j)=(1-p)^2 p^j$. For the case $p\to 1$, this, however, converges to zero, not $1-e^{-j}$.
Where is my mistake? Or is this task truly fundamentally flawed?
 A: Your argument was fundamentally flawed. First of all, $\newcommand{\P}{\Bbb P}\P[Y=k]\neq (1-p)^2p^k$. Instead,
$$
\P[Y=k]=\P[(1-p)X=k]=\P\left[{X}=\frac{k}{1-p}\right]=(1-p)p^{k/(1-p)}
$$
In particular, while $X$ is supported on the nonnegative integers, $Y$ is supported on the reals of the form $(1-p)k$ for $k\in \{0,1,2,\dots\}$.
Secondly, and more importantly, to prove distributional convergence, you do not look at the limit of the probability mass functions. For example, the central limit theorem implies that if $X\sim \operatorname{Binomial}(n,\frac12)$, then $\frac{X-n/2}{\sqrt{n}/2}$ converges in distribution to a standard normal random variable $Z$. However, for any irrational number, like $\pi$, we have $P(X=\pi)=0$ always, yet the limit probability density function is nonzero at $\pi$. This is the problem with comparing discrete distributions with continuous ones.
To fix this, you need to compare the cumulative distributions of $Y$ and the exponential distribution. That is, you need to show that for any $x\in \mathbb R^+$, that
$$
\lim_{p\to 1} P(Y<x)=1-e^{-x}
$$
The event $Y<x$ will be a certain finite summation, and you will have to show that summation approaches an integral which will evaluate to $1-e^{-x}$.
A: One way of proving this is to use Laplace transforms. The Laplace transform of an $exp(1)$ r.v. $Y$ is $Ee^{-sY}=\frac 1  {1+s}$.
Now $Ee^{-sY_p}=\sum e^{-s(1-p)k} (1-p)p^{k}$. This is a geometric sum and you get $Ee^{-sY_p}=\frac {1-p} {1-pe^{-s(1-p)}}$. An application of L'Hopital's Rule shows that the limit of this as $ p \to 1$ is $\frac 1 {1+s}$, as required.
A: Let's use characteristic functions
$$E[e^{itY}]=E[e^{it(1-p)X}]=\sum_{k=0}^{\infty}(1-p)p^{k}e^{itk(1-p)}=\frac{1-p}{1-pe^{it(1-p)}}$$ . Using geometric sum formula.
Now as $p\to 1$ it is $\frac{0}{0}$ form.
So using L'hospital.
$$\lim_{p\to 1}\frac{1-p}{1-pe^{it(1-p)}}= \frac{1}{1-it}=\phi_{Z}(t)=E[e^{itZ}]$$.
Where $\phi_{Z}(t)$ denotes the characteristic function of a $Z$ which is a  $\text{exponential}(1)$ variate. So in particular the cdf of $Z$ will be that of a $\text{exponential}(1)$ variate.
Hence as $p\to 1$ . The cdf of $Y$ will converge pointwise to cdf of $Z$.
In particular we are using the fact that if $X_{n}\to X$ in distribution then $\phi_{X_{n}}(t)\to \phi_{X}(t)$ as $n\to\infty$ and conversely there exist a random variable $Z$ with same distribution as $X$ such that $X_{n}\to Z$ in distribution.
Also following standard notation , $X_{p}$ is $Geo(1-p)$ and not $Geo(p)$ distributed.
