Question

For every n>1 the random variable Xn is exponential with parameter λn, where λn→λ>0 an X is an exponential with parameter λ, then show that Xn converges in distribution to X

Answer 1

Answer on Question #59972 – Math – Statistics and Probability

Question

For every $n > 1$ the random variable $X_n$ is exponential with parameter $\lambda_n$, where $\lambda_n \to \lambda > 0$ and $X$ is an exponential with parameter $\lambda$, then show that $X_n$ converges in distribution to $X$.

Solution

We say that $\{X_n\}$ converges in distribution to the random variable $X$ if

$$\lim_{n \to \infty} F_n(t) = F(t),$$

at every value $t$ where $F$ is continuous cumulative distribution function.

$$F_n(t) = 1 - e^{-\lambda_n t}, \text{ for } t \geq 0$$ $$F(t) = 1 - e^{-\lambda t}, \text{ for } t \geq 0$$ $$\lim_{n \to \infty} F_n(t) = \lim_{n \to \infty} \left(1 - e^{-\lambda_n t}\right) = 1 - \lim_{n \to \infty} \left(e^{-\lambda_n t}\right) = 1 - \exp\left(-t \lim_{n \to \infty} \lambda_n\right) = 1 - \exp(-t\lambda) = F(t),$$

for $t \geq 0$.

Thus, $\{X_n\}$ converges in distribution to the random variable $X$.

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