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.

Question
Normal distributions
asked 2021-02-09
Assume two normal distributions where \(\displaystyle\mu_{{1}}={0.0001},\sigma_{{1}}={0.01},\mu_{{2}}=-{.0002},\sigma_{{2}}={0.015},{\quad\text{and}\quad}ρ={.45}\). Using \(\displaystyle{z}_{{s}}={\left\lbrace{.814},{.259}\right\rbrace}\), generate \(\displaystyle{z}_{{1}}{\quad\text{and}\quad}{z}_{{2}}\).

Answers (1)

2021-02-10
Step 1
Given information-
\(\displaystyle\mu_{{1}}={0.0001},\sigma_{{1}}={0.01},\mu_{{2}}=-{.0002},\sigma_{{2}}={0.015},{\quad\text{and}\quad}ρ={.45}\).
Using \(\displaystyle{z}_{{s}}={\left\lbrace{0.814},{0.259}\right\rbrace}\)
So, X = 0.814, Y = 0.259
Step 2
So,
\(\displaystyle{z}_{{1}}=\frac{{{X}-\mu_{{1}}}}{\sigma_{{1}}}\)
\(\displaystyle{z}_{{1}}=\frac{{{0.814}-{0.0001}}}{{0.01}}={81.39}\)
\(\displaystyle{z}_{{2}}=\frac{{\frac{{{Y}-\mu_{{2}}}}{\sigma_{{2}}}-{p}{z}_{{1}}}}{{\sqrt{{{1}-{p}^{{2}}}}}}\)
\(\displaystyle{z}_{{2}}={\left(\frac{{{0.259}-{\left(-{0.0002}\right)}}}{{0.015}}-{0.45}\times{81.39}\right)}={21.6628}\)
So, the value of, \(\displaystyle{z}_{{1}}={81.39}{\quad\text{and}\quad}{z}_{{2}}=-{21.6628}\).
0

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