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?
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?
Show 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.
Updated On: Aug 3, 2026
- \(V\) has normal distribution with mean \(5\) and variance \(25\)
- \(V\) has chi-square distribution with \(5\) degrees of freedom
- \(V\) has exponential distribution with mean \(5\)
- \(e^{-V/5}\) has uniform distribution over \((0,1)\)
Show Solution
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The Correct Option is C, D
Solution and Explanation
Step 1: Identify the distribution of X_n.
\(X_n\) is a geometric random variable on \(\{0,1,2,\ldots\}\) with success probability \(p_n=\dfrac{1}{5n}\), since \(P(X_n=x)=p_n(1-p_n)^x\). Its survival function is
\[ P(X_n\geq m)=(1-p_n)^m,\qquad m=0,1,2,\ldots \]
Step 2: Find the limiting survival function of Z_n.
For \(t\geq0\),
\[ P(Z_n>t)=P(X_n>nt)\approx(1-p_n)^{nt}=\left(1-\frac{1}{5n}\right)^{nt}. \]
As \(n\to\infty\), using the standard limit \(\left(1-\dfrac{a}{n}\right)^{n}\to e^{-a}\), we get
\[ \left(1-\frac{1}{5n}\right)^{nt}\to e^{-t/5}. \]
So \(P(V>t)=e^{-t/5}\), which is the survival function of an Exponential distribution with rate \(1/5\), that is mean \(5\).
Step 3: Check (A) and (B).
V is exponential, not normal and not chi-square, so (A) and (B) are FALSE.
Step 4: Check (C).
We found \(V\sim\text{Exponential}(\text{mean}=5)\), so (C) is TRUE.
Step 5: Check (D).
The CDF of V is \(F_V(v)=1-e^{-v/5}\). By the probability integral transform, \(F_V(V)=1-e^{-V/5}\sim\text{Uniform}(0,1)\). Since \(U\sim\text{Uniform}(0,1)\) implies \(1-U\sim\text{Uniform}(0,1)\) as well, \(e^{-V/5}\sim\text{Uniform}(0,1)\) too. So (D) is TRUE.
Final Answer:
Z_n converges to an exponential distribution with mean 5, and its transform e^(-V/5) is uniform on (0,1). \[ \boxed{\text{(C) and (D)}} \]
\(X_n\) is a geometric random variable on \(\{0,1,2,\ldots\}\) with success probability \(p_n=\dfrac{1}{5n}\), since \(P(X_n=x)=p_n(1-p_n)^x\). Its survival function is
\[ P(X_n\geq m)=(1-p_n)^m,\qquad m=0,1,2,\ldots \]
Step 2: Find the limiting survival function of Z_n.
For \(t\geq0\),
\[ P(Z_n>t)=P(X_n>nt)\approx(1-p_n)^{nt}=\left(1-\frac{1}{5n}\right)^{nt}. \]
As \(n\to\infty\), using the standard limit \(\left(1-\dfrac{a}{n}\right)^{n}\to e^{-a}\), we get
\[ \left(1-\frac{1}{5n}\right)^{nt}\to e^{-t/5}. \]
So \(P(V>t)=e^{-t/5}\), which is the survival function of an Exponential distribution with rate \(1/5\), that is mean \(5\).
Step 3: Check (A) and (B).
V is exponential, not normal and not chi-square, so (A) and (B) are FALSE.
Step 4: Check (C).
We found \(V\sim\text{Exponential}(\text{mean}=5)\), so (C) is TRUE.
Step 5: Check (D).
The CDF of V is \(F_V(v)=1-e^{-v/5}\). By the probability integral transform, \(F_V(V)=1-e^{-V/5}\sim\text{Uniform}(0,1)\). Since \(U\sim\text{Uniform}(0,1)\) implies \(1-U\sim\text{Uniform}(0,1)\) as well, \(e^{-V/5}\sim\text{Uniform}(0,1)\) too. So (D) is TRUE.
Final Answer:
Z_n converges to an exponential distribution with mean 5, and its transform e^(-V/5) is uniform on (0,1). \[ \boxed{\text{(C) and (D)}} \]
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