Let $X_1,X_2$ be i.i.d. $N(0,1)$ and $U_1,U_2$ be i.i.d. $U(0,1)$ and independent of $X_1,X_2$. Define $$Z_1=\frac{(X_{1}U_{1}+X_{2}U_{2})}{\sqrt{U_{1}^2+U_{2}^2}}.$$ Find the distribution of $Z$. Answer: $$Z_1=\frac{(X_{1}U_{1}+X_{2}U_{2})}{\sqrt{U_{1}^2+U_{2}^2}}\stackrel{d}{=}\frac{(X_{1}+X_{2})U_{1}}{\sqrt{2U_{1}^2}}\stackrel{d}{=}\frac{X_{1}+X_{2}}{\sqrt2}\sim N(0,1).$$

But if we proceed one step further, then $\frac{X_{1}+X_{2}}{\sqrt2}\stackrel{d}{=}\frac{2X_{1}}{\sqrt2}\sim N(0,2).$

Which is correct?

  • 3
    $\begingroup$ It's clearly not the case that $X_1+X_2$ has the same distribution as $X_1,$ with few exceptions. You see this easily by considering the simplest possible random variables such as Bernoulli$(1/2)$ variables. Now $X_1+X_1$ cannot even attain the value $1,$ even though that has a 50% chance of occurring for $X_1+X_2$! $\endgroup$ – whuber Sep 13 '18 at 19:39

It appears $Z\sim N(0,1)$. See first figure below.

Further, $X_1+X_2 \stackrel{d}{\ne} 2X_1$, where $\stackrel{d}{\ne}$ means not equal in distribution. See second figure below.

Empirical Results

Added figure: $X_1+X_2 \stackrel{d}{\ne} 2X_1$.


% MATLAB R2018a 
n = 50000;
U1 = rand(n,1);
U2 = rand(n,1);
X1 = normrnd(0,1,n,1);
X2 = normrnd(0,1,n,1);
Z = ((X1.*U1)+(X2.*U2))./sqrt((U1.^2)+(U2.^2));   

pd1 = makedist('Normal',0,1);
pd2 = makedist('Normal',0,2);

figure, hold on, box on
title('Empirical Distribution of Z (50000 samples)')
XAxis = get(gca,'XTick');
XSupport = -5:0.1:5;
xlim([-5 5])

% Edit: Added second test
    Y = X1+X2;
    W = 2*X1;
   figure, hold on, box on
   title('Empirical Distribution of (X_1+X_2) and 2X_1 (50000 samples)')
   XAxis = get(gca,'XTick');
   XSupport = -8:0.1:8;
   xlim([-8 8])
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  • 2
    $\begingroup$ Re "It appears" at the beginning: this result is exact. Provided $(V_1,V_2)$ is any bivariate random variable independent of $(X_1,X_2)$ and assuming $\Pr((V_1,V_2)=(0,0))=0,$ then $Z = (X_1V_1+X_2V_2)/\sqrt{V_1^2+V_2^2}$ will have a standard Normal distribution. The reason is that for every possible value of $(V_1,V_2),$ the linear combination $X_1V_1+X_2V_2$ (obviously) has a zero-mean Normal distribution with variance $V_1^2+V_2^2.$ Thus the conditional distribution of $Z$ given $(V_1,V_2)$ is almost surely standard Normal, whence the marginal distribution of $Z$ must be standard Normal. $\endgroup$ – whuber Sep 13 '18 at 21:06
  • 1
    $\begingroup$ Thanks @whuber. I didn't want to say that without also having the theoretical argument for someone to follow. $\endgroup$ – SecretAgentMan Sep 13 '18 at 21:11
  • 2
    $\begingroup$ I figured as much, which is why I thought the comment might be helpful. BTW, right-clicking on $\TeX$ rendered on these pages gives you a way to see the underlying markup. That's a quick and easy way to learn how a particular symbol was typeset in the question (and you can copy and paste it into your answer to save time and limit typographical errors). $\endgroup$ – whuber Sep 13 '18 at 21:14

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