Question

Let \(\{X_n\}_{n\geq1}\) be a sequence of random variables having the following probability mass function \[ P(X_n=x)=\frac{1}{5n}\left(1-\frac{1}{5n}\right)^{x},\quad x=0,1,2,\ldots;\ n\in\mathbb{N}. \] Define \(Z_n=\dfrac{X_n}{n}\), \(n\in\mathbb{N}\), and let \(V\) be a random variable. If \(Z_n\xrightarrow{d}V\), as \(n\to\infty\), then which of the following statements is/are correct?

• A. \(V\) has normal distribution with mean \(5\) and variance \(25\)
• B. \(V\) has chi-square distribution with \(5\) degrees of freedom
• C. \(V\) has exponential distribution with mean \(5\)
• D. \(e^{-V/5}\) has uniform distribution over \((0,1)\)

Hint:

Find the limiting survival function P(Zn greater than t) as n grows large; it should converge to e^(-t/5), the survival function of an exponential distribution with mean 5.

Answer 1

The Correct Option is C, D

Step 1: Write the moment generating function of X_n.
For a geometric random variable on $\{0,1,2,\ldots\}$ with parameter $p_n$, the MGF is
\[ M_{X_n}(s)=E(e^{sX_n})=\frac{p_n}{1-(1-p_n)e^{s}},\qquad e^s(1-p_n)<1. \]

Step 2: Get the MGF of Z_n=X_n/n.
\[ M_{Z_n}(s)=E(e^{sX_n/n})=\frac{p_n}{1-(1-p_n)e^{s/n}}, \]
with $p_n=\frac{1}{5n}$.

Step 3: Take the limit as n grows large.
For large n, $e^{s/n}\approx1+\frac{s}{n}$ and $1-p_n\approx1-\frac{1}{5n}$, so
\[ (1-p_n)e^{s/n}\approx1+\frac{s-1/5}{n}. \]
So the denominator $1-(1-p_n)e^{s/n}\approx\frac{1/5-s}{n}$, and the numerator $p_n=\frac{1}{5n}$. Dividing the two,
\[ M_{Z_n}(s)\to\frac{1/5}{1/5-s}=\frac{1}{1-5s}, \]
which is the MGF of an Exponential distribution with mean 5, valid for $s<1/5$.

Step 4: Match with the options.
So $V\sim$ Exponential(mean 5), confirming (C) and ruling out (A) normal and (B) chi-square.

Step 5: Apply the probability integral transform.
For any continuous random variable, plugging it into its own CDF gives a Uniform(0,1) variable. Here $F_V(v)=1-e^{-v/5}$, so $1-e^{-V/5}\sim U(0,1)$. Since reflecting a Uniform(0,1) variable about one half keeps it Uniform(0,1), $e^{-V/5}=1-(1-e^{-V/5})$ is also $U(0,1)$. So (D) holds.

Final Answer:
The MGF limit confirms V is exponential with mean 5, and (D) follows from the probability integral transform. \[ \boxed{\text{(C) and (D)}} \]