Question

For each $ n \in N=\{1,2, \ldots\} $, let $ X_{n} $ be a random variable with probability distribution given by \[ P\left(X_{n}=j\right)=\frac{2 j}{n(n+1)}, \quad j=1, \ldots, n . \] Obtain: (a) the cumulative distribution function of the random variable $ Y_{n}=\frac{X_{n}}{n} $; (b) the limit, in distribution, of the sequence $ \left(Y_{n}\right)_{n \geq 1} $.

          For each \( n \in N=\{1,2, \ldots\} \), let \( X_{n} \) be a random variable with probability distribution given by
\[
P\left(X_{n}=j\right)=\frac{2 j}{n(n+1)}, \quad j=1, \ldots, n .
\]
Obtain:
(a) the cumulative distribution function of the random variable \( Y_{n}=\frac{X_{n}}{n} \);
(b) the limit, in distribution, of the sequence \( \left(Y_{n}\right)_{n \geq 1} \).

For each n ∈ N={1,2, …}, let Xn be a random variable with probability distribution given by

P(Xn=j)=(2 j)/(n(n+1)), j=1, …, n .

Obtain:
(a) the cumulative distribution function of the random variable Yn=(Xn)/(n);
(b) the limit, in distribution, of the sequence (Yn)n ≥ 1.

Added by Gui G.

Elementary Statistics

Robert Johnson, Patricia Kuby 11th Edition

Chapter 5

Instant Answer

Solved by Expert Madhur L

Step 1

The CDF is defined as $F_{Y_n}(y) = P(Y_n \leq y)$. Since $Y_n = \frac{X_n}{n}$, we have $P(Y_n \leq y) = P\left(\frac{X_n}{n} \leq y\right) = P(X_n \leq ny)$. Now, we can write the CDF as a sum of probabilities: $$F_{Y_n}(y) = P(X_n \leq ny) = Show more…

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For each $n \in \mathbb{N} = \{1, 2, \ldots \}$, let $X_n$ be a random variable with probability distribution given by $$P(X_n = j) = \frac{2j}{n(n + 1)}, \quad j = 1, \ldots, n.$$ Obtain:\\ (a) the cumulative distribution function of the random variable $Y_n = \frac{X_n}{n}$;\\ (b) the limit, in distribution, of the sequence $(Y_n)_{n \geq 1}$.