Tương quan của các biến ngẫu nhiên log-normal

Cho và các biến ngẫu nhiên bình thường có hệ số tương quan , làm cách nào để tìm mối tương quan giữa các biến ngẫu nhiên sau và ? $X_1$ $X_2$ $\rho$ $Y_1$ $Y_2$

$Y_1 = a_1 \exp(\mu_1 T + \sqrt{T}X_1)$

$Y_2 = a_2 \exp(\mu_2 T + \sqrt{T}X_2)$

Now, if $X_1 = \sigma_1 Z_1$ and $X_2 = \sigma_1 Z_2$ , where $Z_1$ and $Z_2$ are standard normals, from the linear transformation property, we get:

$Y_1 = a_1 \exp(\mu_1 T + \sqrt{T}\sigma_1 Z_1)$

$Y_2 = a_2 \exp(\mu_2 T + \sqrt{T}\sigma_2 (\rho Z_1 + \sqrt{1-\rho^2}Z_2)$

Now, how to go from here to compute correlation between $Y_1$ and $Y_2$ ?

correlation random-variable lognormal

— user862
nguồn

@user862, hint: use chracteristic function of bivariate normal.

— mpiktas

See equation (11) in stuart.iit.edu/shared/shared_stuartfaculty/whitepapers/… (but watch out for the awful typesetting).

— whuber

I assume that $X_1\sim N(0,\sigma_1^2)$ and $X_2\sim N(0,\sigma_2^2)$ . Denote $Z_i=\exp(\sqrt{T}X_i)$ . Then

\begin{aligned} \log (Z_{i}) \sim N (0, T σ_{i}^{2}) \end{aligned}

$\begin{align} \log(Z_i)\sim N(0,T\sigma_i^2) \end{align}$ so

Z_{i}

$Z_i$ are log-normal. Thus

\begin{aligned} E Z_{i} & = \exp (\frac{T σ_{i}^{2}}{2}) \\ v a r (Z_{i}) & = (\exp (T σ_{i}^{2}) - 1) \exp (T σ_{i}^{2}) \end{aligned}

$\begin{align} EZ_i&=\exp\left(\frac{T\sigma_i^2}{2}\right)\\ var(Z_i)&=(\exp(T\sigma_i^2)-1)\exp(T\sigma_i^2) \end{align}$ and

\begin{aligned} E Y_{i} & = a_{i} \exp (μ_{i} T) E Z_{i} \\ v a r (Y_{i}) & = a_{i}^{2} \exp (2 μ_{i} T) v a r (Z_{i}) \end{aligned}

$\begin{align} EY_i&=a_i\exp(\mu_iT)EZ_i\\ var(Y_i)&=a_i^2\exp(2\mu_iT)var(Z_i) \end{align}$

Then using the formula for m.g.f of multivariate normal we have

\begin{aligned} E Y_{1} Y_{2} & = a_{1} a_{2} \exp ((μ_{1} + μ_{2}) T) E \exp (\sqrt{T} X_{1} + \sqrt{T} X_{2}) \\ = a_{1} a_{2} \exp ((μ_{1} + μ_{2}) T) \exp (\frac{1}{2} T (σ_{1}^{2} + 2 ρ σ_{1} σ_{2} + σ_{2}^{2})) \end{aligned}

$\begin{align} EY_1Y_2&=a_1a_2\exp((\mu_1+\mu_2)T)E\exp(\sqrt{T}X_1+\sqrt{T}X_2)\\ &=a_1a_2\exp((\mu_1+\mu_2)T)\exp\left(\frac{1}{2}T(\sigma_1^2+2\rho\sigma_1\sigma_2+\sigma_2^2)\right) \end{align}$ So

\begin{aligned} c o v (Y_{1}, Y_{2}) & = E Y_{1} Y_{2} - E Y_{1} E Y_{2} \\ = a_{1} a_{2} \exp ((μ_{1} + μ_{2}) T) \exp (\frac{T}{2} (σ_{1}^{2} + σ_{2}^{2})) (\exp (ρ σ_{1} σ_{2} T) - 1) \end{aligned}

$\begin{align} cov(Y_1,Y_2)&=EY_1Y_2-EY_1EY_2\\ &=a_1a_2\exp((\mu_1+\mu_2)T)\exp\left(\frac{T}{2}(\sigma_1^2+\sigma_2^2)\right)(\exp(\rho\sigma_1\sigma_2T)-1) \end{align}$

Now the correlation of $Y_1$ and $Y_2$ is covariance divided by square roots of variances:

\begin{aligned} ρ_{Y_{1} Y_{2}} = \frac{\exp (ρ σ_{1} σ_{2} T) - 1}{\sqrt{(\exp (σ_{1}^{2} T) - 1) (\exp (σ_{2}^{2} T) - 1)}} \end{aligned}

$\begin{align} \rho_{Y_1Y_2}=\frac{\exp(\rho\sigma_1\sigma_2T)-1}{\sqrt{\left(\exp(\sigma_1^2T)-1\right)\left(\exp(\sigma_2^2T)-1\right)}} \end{align}$

— mpiktas
nguồn

Note that as long as the approximation

e^{x} ≅ 1 + x

$e^x \cong 1+ x$ is valid on the final formula found above one has

ρ_{Y_{1} Y_{2}} ≅ ρ

$\rho_{Y_1Y_2} \cong \rho$ .

— danbarros