A) \[e\]
B) \[2e\]
C) \[{{e}^{2}}\]
D) \[2e\]
Correct Answer: D
Solution :
We know that, \[\sigma _{ax+by}^{2}={{a}^{2}}\sigma _{x}^{2}+{{b}^{2}}\sigma _{x}^{2}+2abr(x,y){{\sigma }_{x}}{{\sigma }_{y}}\] but\[r=0\]is given \[\therefore \] \[\sigma _{ax+by}^{2}={{a}^{2}}\sigma _{x}^{2}+{{b}^{2}}\sigma _{y}^{2}\] ...(i) Putting\[a=b=1\]in Eq. (i), we get \[\sigma _{x+y}^{2}=\sigma _{x}^{2}+\sigma _{y}^{2}\] \[\Rightarrow \] \[\sigma _{u}^{2}=\sigma _{x}^{2}+\sigma _{y}^{2}\] ...(ii) Again putting\[a=1,\text{ }b=-1\]in Eq. (ii), we get \[\Rightarrow \] \[\left. \begin{align} & \sigma _{x-y}^{2}=\sigma _{x}^{2}+\sigma _{y}^{2} \\ & \sigma _{v}^{2}=\sigma _{x}^{2}+\sigma _{y}^{2} \\ \end{align} \right]\] Now, \[r(u,v)=\frac{Cov(u,v)}{{{\sigma }_{u}}{{\sigma }_{v}}}\] \[r(u,v)=\frac{\sigma _{x}^{2}-\sigma _{y}^{2}}{\sigma _{u}^{2}}\] \[(\because Cov(u,v)=\sigma _{x}^{2}-\sigma _{y}^{2}and\,{{\sigma }_{u}}={{\sigma }_{v}})\] \[\Rightarrow \] \[r(u,v)=\frac{\sigma _{x}^{2}-\sigma _{y}^{2}}{\sigma _{x}^{2}+\sigma _{y}^{2}}\] [using Eq. (ii)]
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