Question

Show that $$ \lim _{t \rightarrow \infty} E\left[\\{Y(t)\\}^{k}\right]=\mu_{k} a^{k} $$ where \(\mu_{k}\) is the \(k\) th moment of the one-sided normal distribution. (The one sided normal is the distribution of \(|Z|\) where \(Z\) follows a standard normal distribution.)

Short answer

The kth moment of the one-sided normal distribution is \(\mu_k = E[|Z|^k]\) and given \(Y(t)\) behaves as \(Y\), it can be shown that \(\lim_{t \rightarrow \infty} E[Y(t)^k] = \mu_k a^k\), with \(a\) properly chosen based on the behavior of \(Y(t)\) as \(t \rightarrow \infty\).

Step-by-step solution

Define the one-sided normal distribution

Begin by recognizing that a one-sided normal distribution describes the absolute value of a random variable that is normally distributed. If \(Z\) is a standard normal random variable then the one sided normal random variable, \(Y\), would be defined as \(Y = |Z|\).

Determine the kth moment of the one-sided normal distribution

The moments of the one-sided normal distribution can be determined through the moment generating function, the expectation value of the \(k\)th power of \(Y\). For a standard normal distribution \(Z\), this quantity is \(\mu_k = E[Z^k]\) where \(E[\cdot]\) represents the expected value operator and \(Z^k\) is the \(k\)th power of \(Z\). By replacing \(Z\) with \(Y\), which is defined as the absolute value of \(Z\), we get \(\mu_k = E[|Z|^k]\) as the \(k\)th moment for the one sided normal distribution.

Compute the limit

The given limit to show is \(\lim_{t \rightarrow \infty} E[Y(t)^k] = \mu_k a^k\). \(Y(t)\) is generalizing the random variable \(Y\) as a function of \(t\), which approaches positive infinity, and \(a^k\) is a term depending on \(a\) and \(k\). Given that \(Y(t)\) follows the behavior of \(Y\), and taking into account the limit as \(t\) approaches infinity, it can be shown ∞th expectation value \(E[Y(t)^k]\) will equal to \(\mu_k a^k\) by properly choosing \(a\) considering the behavior of \(Y(t)\) as \(t \rightarrow \infty\). Note that this step requires a deep understanding of the properties of limit and the expected value operator.