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I would like to generate one random numer $single\sim N(0,1)$ and create the vector that contains only this one number: $one = [single, single, ..., single]$ . Later, I would like to combine with vector of multiple random numbers $several \sim N(0,1)$ using $weight$:

$New=weight*one+\sqrt(1-weight^2)*several$

Is it $New\sim N(0,1)?$

I thought so, because both $one$, and $several$ are random and generated from normal distribution. However, right now, I am not so sure. $one$ is in fact a vector of the same values so it might be treateat as constant here.

I tried to use R code to check below:

results<-matrix(ncol=4,nrow<-0)

colnames(results)<-c("sd(Normal)", "mean(Normal)", "sd(New)", "mean(new)")

for(i in 1:100000){

  set.seed(i)
  one<-rnorm(1,mean=0,sd=1)
  several<-rnorm(4000, mean=0, sd=1)

  weight<-0.20

  NORMAL<-several #this is N(0,1)
  NEW<- weight*one+sqrt(1-weight^2)*several #not sure if N(0,1)

  vector<- c(sd(NORMAL), mean(NORMAL), sd(NEW), mean(NEW))
  results<- rbind(results, vector)
}

colMeans(results)*100 #BP
sqrt(1-weight^2)*100 #BP

Results:

      sd(Normal)   mean(Normal)      sd(New)       mean(new) 
      99.99548850  -0.00423462       97.97516937   0.04152757

The results sd(New) are very similar to the $sqrt(1-weight^2)*100=97.97959$ and that raised my question about $New$ variable.

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  • 3
    $\begingroup$ Mathematically, your question doesn't make sense until you can explain what it means to add a scalar to a vector. Could you edit the post to clarify this point? $\endgroup$ – whuber Aug 9 '19 at 12:04
  • $\begingroup$ Thanks for your comment: I have edited to make sure we are adding vectors. $\endgroup$ – Lohengrin Aug 9 '19 at 12:44
  • $\begingroup$ In symbols, are you asking about the distribution of the vector $(Y_i)$ where $Y_i=a X + b Z_i$ with $a,b$ constants with $a^2+b^2=1,$ $X$ Normally distributed, $Z_i$ multinormally distributed independently of $X$? $\endgroup$ – whuber Aug 9 '19 at 12:51
  • $\begingroup$ Yes, in R set.seed(1) we have: $X=[0.7083495, 0.7083495, 0.7083495, ...], Z_i = [-0.03846413, 0.94647475, 0.84385583, ... ]$ wheras $ Y_i = [0.1039829,1.0690220, 0.9684764]$ I am not sure if $Y_i ~N(0,1)$ $\endgroup$ – Lohengrin Aug 9 '19 at 13:28
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$New=weight*one+\sqrt(1-weight^2)*several$ change to $$Y_{ij}=wX_i +\sqrt{1-w^2}Z_{ij}$$ where $X,Z$ follows standard normal distribution and independent from each other, and $ 0\le w\le 1$.

Let $\gamma_i = wX_i$, then $\gamma_i \sim N(0,w^2)$. Let $\epsilon_{ij} = \sqrt{1-w^2}Z_{ij}$ then $\epsilon \sim N(0, 1-w^2)$. $$Y_{ij}=\gamma_i+\epsilon_{ij}$$ It is the random part in the mixed model. Let see the whole picture of $Y_{ij}$. Let $i=1,...,I$ and $j=1,..., J$. Let $Y = (Y_{11} ,Y_{12},...,Y_{1J},...,Y_{i1},...,Y_{IJ})'$ $$\begin{pmatrix} Y_{11}\\ Y_{12}\\ ...\\ Y_{1J} \end{pmatrix} = \begin{pmatrix} 1& 1 &0 &...&0\\ 1& 0& 1 &...&0\\ ...&...&...&...&...\\ 1&0&0&...&1 \end{pmatrix} \begin{pmatrix} \gamma_1\\ \epsilon_{11}\\ \epsilon_{12}\\ ...\\ \epsilon_{1J} \end{pmatrix}$$ $$Var\begin{pmatrix} \gamma_1\\ \epsilon_{11}\\ \epsilon_{12}\\ ...\\ \epsilon_{1J} \end{pmatrix} = \begin{pmatrix} w^2& 0 & ...&0\\ 0&1-w^2&...&0\\ ...&...&...&...\\ 0&0&...&1-w^2 \end{pmatrix}$$ Following $Var(AX)=AVar(X)A'$, you can get $$Var\begin{pmatrix} Y_{11}\\ Y_{12}\\ ...\\ Y_{1J} \end{pmatrix} = \begin{pmatrix} 1& 1 &0 &...&0\\ 1& 0& 1 &...&0\\ ...&...&...&...&...\\ 1&0&0&...&1 \end{pmatrix} \begin{pmatrix} w^2& 0 & ...&0\\ 0&1-w^2&...&0\\ ...&...&...&...\\ 0&0&...&1-w^2 \end{pmatrix}\begin{pmatrix} 1& 1 &0 &...&0\\ 1& 0& 1 &...&0\\ ...&...&...&...&...\\ 1&0&0&...&1 \end{pmatrix}=\begin{pmatrix} 1&w^2&...&w^2\\ w^2& 1&...&w^2\\ ...&...&...&...\\ w^2&w^2&...&1 \end{pmatrix}_{J\times J} =\Sigma$$

Then $Y\sim N(0, I\otimes \Sigma)$.

What you did is setting $I=1$ and $J=4000$. The sample variance of $Y_{1j}, j=1,...,4000$ is the estimate of conditional variance $Var(Y_{ij}|i=1) = Var(Y_{1j}|\gamma_1) = Var(\epsilon_{1j}) = 1-w^2$

If you want to verify the unconditional variance of $Y_{ij}$ being 1, you can set $I = 200$ and $J=20$. and re-run roue code.

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  • $\begingroup$ Many thanks for your help - I see that you put a lot of effort into writing the answer. I just want to clarify this point: $\Sigma = \begin{pmatrix} 1&w^2&...&w^2\\ w^2& 1&...&w^2\\ ...&...&...&...\\ w^2&w^2&...&1 \end{pmatrix}_{I\times I}$ how did you get this one? $\endgroup$ – Lohengrin Aug 10 '19 at 16:10
  • $\begingroup$ The details were added in the Answer. Need matrix skill. $\endgroup$ – user158565 Aug 10 '19 at 17:54
  • $\begingroup$ All is clear now, many many thanks user158565! $\endgroup$ – Lohengrin Aug 12 '19 at 21:19

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