# Linear transformation from one sample to another

Generate a Sample $$\underline{Z_1}$$ $$\underline{Z_2}\dots \underline{Z_{5000}}$$ , while $$\underline{Z_i} \sim N_2[(0,0)^T,I_2]$$

generate new sample with size of $$n = 5000$$ by applying linear transformation on $$\underline{Z_i}$$

$$\underline{X_1}$$ $$\underline{X_2}\dots \underline{X_{5000}}$$ , while $$\underline{X_i} \sim N_2[(1,2)^T,\begin{pmatrix}2&1.5\\ 1.5&2\end{pmatrix}]$$.

My attempt:

n1 <- 5000
mu <- c(0,0)
sigma <- diag(2)
y2 <- mvrnorm(n1,mu,sigma)


I have generated the first sample but aside from that , I have no idea how to continue..

Update $$X = A*Z + \mu$$ while $$\mu = (1,2)^T.$$

To find $$A$$ we use this equation $$\Sigma = A*A^T$$.

ed = eigen(Sigma)
A = ed$vectors %*% diag(sqrt(ed$values))


but still I'm not getting the right value for $$A$$.

because $$A*A^T\neq \Sigma$$.

## 1 Answer

Say we want to sample $$x$$ from $$N(a,b)$$. We could certainly do this in R using rnorm(a,b). However, we could also sample $$z$$ from $$N(0,1)$$ and apply the linear transformation $$a + zb$$. This transformation would then give us samples from $$N(a,b)$$.

What I've described is for the univariate case. You can easily extend this to the multivariate case.

• Thanks , this is helpful for my case , but for that I need to find $A*A^T=\Sigma$ ,please check the updated question. – Mahajna Dec 20 '20 at 10:41
• I think you're overthinking this. Recall that $AI = A$, where $I$ is the identity matrix. Given this, what do you think $A$ should be? – ralph Dec 20 '20 at 20:59