Show that $Y_1 X_1 + Y_2 X_2$ $\,{\buildrel d \over =}\,$ $(Y_1^2+Y_2^2)^{1/2}X_1$ I would like verification of my solution to the following problem.
QUESTION:
Let $X_1, X_2 \,{\buildrel iid \over \sim }\, N(0,1) $  and let $Y_1, Y_2$ be two independent random variables 
($X_1, X_2, Y_1, Y_2$ all independent)
Show that $Y_1 X_1 + Y_2 X_2$ $\,{\buildrel d \over =}\,$ $(Y_1^2+Y_2^2)^{1/2}X_1$

My solution
I went with using the CF.
For the LEFT HAND SIDE:
$\phi_{LHS}(t)=\phi_{Y_1 X_1 + Y_2 X_2}(t) = E[e^{it (Y_1 X_1 + Y_2 X_2)}] \,{\buildrel ind \over =}\, E[e^{it (Y_1 X_1)}] E[e^{it (Y_2 X_2)}]$
Using conditional expectation, for each $j, E[e^{it Y_j X_j}]=E(E[e^{it Y_j X_j}|Y_j]) $
Now,$E[e^{it Y_j X_j}|Y_j=y]=E[e^{it y X_j}] = \phi_{X_j}(ty)=\exp(-t^2y^2/2 )$ because the characteristic function for a standard normal RV Z is $\phi_Z(t)=\exp(-t^2/2)$
So, $E(E[e^{it Y_j X_j}|Y_j])=E(e^{-t^2Y_j^2/2} )$
Thus, $E[e^{it (Y_1 X_1)}] E[e^{it (Y_2 X_2)}] =E[e^{-t^2Y_1^2/2}] E[e^{-t^2Y_2^2/2}]$ $$\,{\buildrel iid \over =}\,  E[e^{-t^2(Y_1^2+Y_2^2)/2}]$$
NOW, For the RIGHT HAND SIDE:
$\phi_{RHS}(t) =\phi_{(Y_1^2+Y_2^2)^{1/2}X_1}(t) = E[e^{it (Y_1^2+Y_2^2)^{1/2}X_1}]$
Applying conditional expectation (as above), $ E[e^{it (Y_1^2+Y_2^2)^{1/2}X_1}] = E(E[e^{it (Y_1^2+Y_2^2)^{1/2}X_1}|(Y_1^2+Y_2^2)^{1/2}])$ $$= E[e^{-t^2(Y_1^2+Y_2^2)/2}]$$
We can see that $\phi_{LHS}(t)=\phi_{RHS}(t)$
THEREFORE, by uniqueness of characteristic functions, $$Y_1 X_1 + Y_2 X_2 \,{\buildrel d \over =}\, (Y_1^2+Y_2^2)^{1/2}X_1.$$
 A: There is more : $Y_1 X_1 + Y_2 X_2 \,{\buildrel d \over =}\, (Y_1^2+Y_2^2)^{1/2}X_1$ conditionally on $(Y_1,Y_2)$.
Indeed, "the" conditional distribution of $Y_1 X_1 + Y_2 X_2$ given $(Y_1=y_1,Y_2=y_2)$ is the distribution of $y_1 X_1 + y_2 X_2$ by independence. And $y_1 X_1 + y_2 X_2 \sim {\cal N}(0, y_1^2+y_2^2)$ by well-known properties of independent Gaussian random variables.
Similarly, the conditional distribution of $(Y_1^2+Y_2^2)^{1/2}X_1$ given $(Y_1=y_1,Y_2=y_2)$ is the distribution of $(y_1^2+y_2^2)^{1/2}X_1 \sim {\cal N}(0, y_1^2+y_2^2)$.
Consequently, the random vectors $(Y_1,Y_2,Y_1 X_1 + Y_2 X_2)$ and $(Y_1,Y_2,(Y_1^2+Y_2^2)^{1/2}X_1)$ have the same distribution.
A: I went with using the CF.
For the LEFT HAND SIDE:
$\phi_{LHS}(t)=\phi_{Y_1 X_1 + Y_2 X_2}(t) = E[e^{it (Y_1 X_1 + Y_2 X_2)}] \,{\buildrel ind \over =}\, E[e^{it (Y_1 X_1)}] E[e^{it (Y_2 X_2)}]$
Using conditional expectation, for each $j, E[e^{it Y_j X_j}]=E(E[e^{it Y_j X_j}|Y_j]) $
Now,$E[e^{it Y_j X_j}|Y_j=y]=E[e^{it y X_j}] = \phi_{X_j}(ty)=\exp(-t^2y^2/2 )$ because the characteristic function for a standard normal RV Z is $\phi_Z(t)=\exp(-t^2/2)$
So, $E(E[e^{it Y_j X_j}|Y_j])=E(e^{-t^2Y_j^2/2} )$
Thus, $E[e^{it (Y_1 X_1)}] E[e^{it (Y_2 X_2)}] =E[e^{-t^2Y_1^2/2}] E[e^{-t^2Y_2^2/2}]$ $$\,{\buildrel iid \over =}\,  E[e^{-t^2(Y_1^2+Y_2^2)/2}]$$
NOW, For the RIGHT HAND SIDE:
$\phi_{RHS}(t) =\phi_{(Y_1^2+Y_2^2)^{1/2}X_1}(t) = E[e^{it (Y_1^2+Y_2^2)^{1/2}X_1}]$
Applying conditional expectation (as above), $ E[e^{it (Y_1^2+Y_2^2)^{1/2}X_1}] = E(E[e^{it (Y_1^2+Y_2^2)^{1/2}X_1}|(Y_1^2+Y_2^2)^{1/2}])$ $$= E[e^{-t^2(Y_1^2+Y_2^2)/2}]$$
We can see that $\phi_{LHS}(t)=\phi_{RHS}(t)$
THEREFORE, by uniqueness of characteristic functions, $$Y_1 X_1 + Y_2 X_2 \,{\buildrel d \over =}\, (Y_1^2+Y_2^2)^{1/2}X_1$$
