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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Statistics

NDA 2016 · Mathematics · Statistics

Two variates, x and y , are uncorrelated and have standard...

Two variates, x and y , are uncorrelated and have standard deviations σx\sigma_x and σy\sigma_y respectively. What is the correlation coefficient between x+y\mathrm{x}+\mathrm{y} and x−y\mathrm{x}-\mathrm{y} ?

Options

  1. Aσxσyσx2+σy2\frac{\sigma_x \sigma_y}{\sigma_x^2+\sigma_y^2}
  2. Bσx+σy2σxσy\frac{\sigma_x+\sigma_y}{2 \sigma_x \sigma_y}
  3. Cσx2−σy2σx2+σy2\frac{\sigma_x^2-\sigma_y^2}{\sigma_x^2+\sigma_y^2}
  4. Dσy−σxσxσy\frac{\sigma_{\mathrm{y}}-\sigma_{\mathrm{x}}}{\sigma_{\mathrm{x}} \sigma_{\mathrm{y}}}

Correct answer

C. σx2−σy2σx2+σy2\frac{\sigma_x^2-\sigma_y^2}{\sigma_x^2+\sigma_y^2}

Step-by-step solution

Let u=(x+y);v=(x−y)u=(x+y) ; v=(x-y)∴uˉ=(xˉ+yˉ);vˉ=(xˉ−yˉ)cov⁡(u,v)=E{(u−uˉ)(v−vˉ)}=E{(x−xˉ)+(y−yˉ)}⋅{(x−xˉ)−(y−yˉ)}=E{(x−xˉ)2−(y−yˉ)2}=σx2−σy2var⁡(u)=E(u−uˉ)2=E{(x−xˉ)+(y−yˉ)}2=σx2+σy2\begin{aligned} & \therefore \bar{u}=(\bar{x}+\bar{y}) ; \bar{v}=(\bar{x}-\bar{y}) \\ & \operatorname{cov}(u, v)=E\{(u-\bar{u})(v-\bar{v})\} \\ & =E\{(x-\bar{x})+(y-\bar{y})\} \cdot\{(x-\bar{x})-(y-\bar{y})\} \\ & =E\left\{(x-\bar{x})^2-(y-\bar{y})^2\right\}=\sigma_x^2-\sigma_y^2 \\ & \operatorname{var}(u)=E(u-\bar{u})^2=E\{(x-\bar{x})+(y-\bar{y})\}^2=\sigma_x^2+\sigma_y^2 \end{aligned}$ThereforeThereforex and yareuncorrelated.are uncorrelated.E(x-\bar{x})(y-\bar{y})=0Similarly,Similarly,\operatorname{var}(v)=\sigma_x^2+\sigma_y^2Thus,Thus,r(u, v)=\frac{\operatorname{cov}(u, v)}{\sqrt{\operatorname{var}(u) \cdot \operatorname{var}(v)}}=\frac{\sigma_x^2-\sigma_y^2}{\sigma_x^2+\sigma_y^2}$
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