Let X_(n)sim Exponential(n) . Show that X_(n)arrow 0 , that is, the sequence x_(1),x_(2)- comwen, o probability to zero random variable X. (4 marks)

To show that the sequence $X_n$ converges to 0, we need to demonstrate that for any positive value $\epsilon$, the probability of $|X_n| > \epsilon$ approaches 0 as $n$ approaches infinity.<br /><br />Given that $X_n \sim Exponential(n)$, we know that the expected value of $X_n$ is $E(X_n) = n$ and the variance is $Var(X_n) = n^2$.<br /><br />Now, let's consider the probability of $|X_n| > \epsilon$. This can be expressed as the complement of the probability that $|X_n| \leq \epsilon$, which is $P(|X_n| > \epsilon) = 1 - P(|X_n| \leq \epsilon)$.<br /><br />To find the probability that $|X_nleq \epsilon$, we can use the cumulative distribution function (CDF) of the exponential distribution. The CDF of an exponential random variable $X$ with rate parameter $\lambda$ is given by $F_X(x) = 1 - e^{-\lambda x}$.<br /><br />In our case, the rate parameter of $X_n$ is $n$, so the CDF of $X_n$ is $F_{X_n}(x) = 1 - e^{-nx}$.<br /><br />Now, let's consider the probability that $|X_n| \leq \epsilon$. This can be expressed as the probability that $X_n$ falls within the interval $[-\epsilon, \epsilon]$. Using the CDF, we can calculate this probability as $P(|X_n| \leq \epsilon) = F_{X_n}(\epsilon) - F_{X_n}(-\epsilon)$.<br /><br />Since $X_n$ is a symmetric random variable around 0, we have $F_{X_n}(-\epsilon) = 1 - F_{X_n}(\epsilon)$. Therefore, $P(|X_n| \leq \epsilon) = 2F_{X_n}(\epsilon) - 1$.<br /><br />Now, let's substitute this expression into the probability that $|X_n| > \epsilon$: $P(|X_n| > \epsilon) = 1 - (2F_{X_n}(\epsilon) - 1) = 2F_{X_n}(\epsilon)$.<br /><br />As $n$ approaches infinity, the term $F_{X_n}(\epsilon)$ approaches 1, since the exponential distribution has a tail probability that approaches 0. Therefore, the probability that $|X_n| > \epsilon$ approaches 0.<br /><br />Hence, we have shown that for any positive value $\epsilon$, the probability of $|X_n| > \epsilon$ approaches 0 as $n$ approaches infinity. This implies that the sequence $X_n$ converges to 0 in probability.