if $X$ and $Y$ are Normal random variables with same parameters. is $(aX+bX)=(aX+bY)$? if $X$ and $Y$ are random variables and both are $\sim N(\mu,\sigma^2)$, ie share the same average and variance. Then we defined a new random variable as the linear combination of them as: $Z=(aX+bY)$ is that the same as saying $Z=(aX+bX)$? Is $Z=(aX+bX)$ the same as saying $Z=((a+b)X)$? 
 A: You have not said how $X$ and $Y$ are jointly distributed, but only how they are marginally distributed.
If they are both marginally distributed as $N(\mu,\sigma^2)$ then:


*

*At one extreme you have $X=Y,$ in which case $X+Y= 2X \sim N(2\mu,4\sigma^2).$

*At an opposite extremem you have $Y = 2\mu-X,$ in which case $X+Y\sim N(2\mu,0),$ i.e. the sum is constant.

*Somewhere in between, $X$ and $Y$ are independent. In that case $X+Y\sim N(2\mu, 2\sigma^2).$

*And there are cases where there is a nonzero correlation between $X$ and $Y$ but it's not $+1$ or $-1.$

*There are also cases in which $X$ and $Y$ are not jointly normally distributed (but still have that same marginal distribution). One of those is where $Y = \pm X$ with plus and minus having equal probabilities and being chosen independently of $X$. In that case $X+Y$ is not normally distributed, since $\Pr(X+Y=0) = 1/2.$

A: I'm absolutely sure the answer is no, unless $X$ and $Y$ are exactly the same random variable $(X \equiv Y).$ But @MarcusLuebke raises good points (+1), especially about independence. 
Let $X$ and $Y$ be independent random variables. Consider $\mu =2,\, \sigma=1$ and
$a =3,\,b=2.$ Let $W_1 = 3X + 2Y,\, W_2 = 3Y + 2X,$ and $W_3 = 3X + 2X = 5X.$
Then you can use standard formulas to show that $E(W_1) = E(W_2) = E(W_3) = 10.$
But $Var(W_1) = Var(W_2) = (a^2 + b^2)\sigma^2 = 13,$ while $Var(W_3) =25.$ All of three of the $W_i$ are normal.
Samples simulated to these specifications have the following histograms.
As noted above, the bottom plot (for $W_3$) shows greater variability: the vertical red lines are at $\mu_w \pm \sigma_w$ in each case.

A: I'm pretty sure the answer is no. An easy way to think about this is physically. AX+BX corresponds to the same event happening twice, whereas AX+BY corresponds to two separate, INDEPENDENT events occuring. With AX+BX, you pull the same point out of the distribution (causing an amplification), whereas with AX+BY, you pull different points (probably causing an increase in variance).
However, I'm not an expert, and there might be some weird mathematical symmetry I don't know about, but I would put it at 94.9% confidence that it is different.
