1
$\begingroup$

I am trying to understand a solution to a question: Let $X_1, X_2 ...$ be independent random variables with distribution $Normal(0, 4)$. Let $K$ be the smallest value of $k$ such that $X_k > 4$ (the index of the first $X_k$ that is greater than 4). Find $E(K)$ in terms of the standard Normal cumulative distribution function $\Phi$.

The solution reads that $K - 1 \sim Geometric(p)$, where $p = P(X_1 > 4) = P(X_1/2 > 2) = 1 - \Phi(2)$, which meant that $E(J) = \frac{1}{1 - \Phi(2)}$.

I understand why $K-1$ has a geometric distribution (We keep trying until one of the $X$ values $> 4$). I don't understand though why $p = P(X_1 > 4)$ (why 1 in particular) and why $P(X_1/2 > 2) = 1 - \Phi(2)$.

  • 1
    If $X_1 \sim N(0,4)$, then $X_1/2$ is standard normally distributed.2012-10-15

2 Answers 2

2

We take the value $1$ in particular in order to simplify, as all the $X_j$ have the same law.

$X_1/2$ follow a standard normal law, as its mean is $0$ and its standard deviation is $1$. Now if $N$ is a random variable following normal standard law then $P(X_1>4)=P(X_1/2>2)=P(N>2)=1-P(N\leq 2)=1-\Phi(2).$

1

Generate a sequence of samples of independent $N(0,4)$ random variables until you get one that exceeds $4$. Now, $P(X_1> 4) = 1 - \Phi(2)$ is the same as $P(X_i> 4) = 1 - \Phi(2)$ for all $i$, right? As to why $P(X_1> 4) = P(X_1/2 > 2)$ equals $1-\Phi(2)$, note that $X_1/2$ is a standard normal random variable since $X_1$ has standard deviation $2$ (and mean $0$).