 # Assume two normal distributions where mu_1 = 0.0001, sigma_1 = 0.01, mu_2 = -.0002, sigma_2 = 0.015, and ρ = .45. Using z_s = { .814, .259 }, generate z_1 and z_2. tabita57i 2021-02-09 Answered
Assume two normal distributions where ${\mu }_{1}=0.0001,{\sigma }_{1}=0.01,{\mu }_{2}=-.0002,{\sigma }_{2}=0.015,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\rho =.45$. Using ${z}_{s}=\left\{.814,.259\right\}$, generate ${z}_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{z}_{2}$.
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Step 1
Given information-
${\mu }_{1}=0.0001,{\sigma }_{1}=0.01,{\mu }_{2}=-.0002,{\sigma }_{2}=0.015,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\rho =.45$.
Using ${z}_{s}=\left\{0.814,0.259\right\}$
So, X = 0.814, Y = 0.259
Step 2
So,
${z}_{1}=\frac{X-{\mu }_{1}}{{\sigma }_{1}}$
${z}_{1}=\frac{0.814-0.0001}{0.01}=81.39$
${z}_{2}=\frac{\frac{Y-{\mu }_{2}}{{\sigma }_{2}}-p{z}_{1}}{\sqrt{1-{p}^{2}}}$
${z}_{2}=\left(\frac{0.259-\left(-0.0002\right)}{0.015}-0.45×81.39\right)=21.6628$
So, the value of, ${z}_{1}=81.39\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{z}_{2}=-21.6628$.
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