Step 1 — Restate the setup

We have \(Z\sim N(0,1)\) and \(Y\sim\chi^{2}(2)\), independent. Define \(W = \dfrac{Z}{Y}\). We need \(\operatorname{Cov}(Y,Z)\) and \(\operatorname{Cov}(Y,W)\).

Step 2 — Assumptions & givens

Independence of \(Z\) and \(Y\). Moments: \(\mathbb{E}[Z]=0\); for \(Y\sim\chi^{2}(2)\), \(\mathbb{E}[Y]=2\). Any measurable function of \(Y\) (e.g., \(1/Y\)) is independent of \(Z\).

Step 3 — Method & formulas

Use the definition \(\operatorname{Cov}(A,B)=\mathbb{E}[AB]-\mathbb{E}[A]\mathbb{E}[B]\). For independent variables, \(\operatorname{Cov}(A,B)=0\). If \(B\) involves an independent variable, \(\mathbb{E}[Z\cdot f(Y)]=\mathbb{E}[Z]\mathbb{E}[f(Y)]\).

Step 4 — Derivation

Covariance of \(Y\) and \(Z\): Since \(Y\) and \(Z\) are independent, \(\operatorname{Cov}(Y,Z)=0\).

Covariance of \(Y\) and \(W\): \(\operatorname{Cov}(Y,W)=\mathbb{E}\!\left[Y\cdot\frac{Z}{Y}\right]-\mathbb{E}[Y]\mathbb{E}\!\left[\frac{Z}{Y}\right]=\mathbb{E}[Z]-\mathbb{E}[Y]\;\mathbb{E}\!\left[\frac{Z}{Y}\right]\). Because \(Z\) is independent of \(Y\), \(\mathbb{E}\!\left[\frac{Z}{Y}\right]=\mathbb{E}[Z]\mathbb{E}\!\left[\frac{1}{Y}\right]=0\). Also \(\mathbb{E}[Z]=0\). Hence \(\operatorname{Cov}(Y,W)=0\).

Step 5 — Final answer

\(\displaystyle \operatorname{Cov}(Y,Z)=0,\qquad \operatorname{Cov}(Y,W)=0.\)

82. Covariance Computation

Similar Questions