3
$\begingroup$

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, $$Y_1 X_1 + Y_2 X_2 \,{\buildrel d \over =}\, (Y_1^2+Y_2^2)^{1/2}X_1.$$

$\endgroup$
1
  • $\begingroup$ is this self study??? $\endgroup$
    – Xi'an
    Mar 23, 2015 at 18:55

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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, $$Y_1 X_1 + Y_2 X_2 \,{\buildrel d \over =}\, (Y_1^2+Y_2^2)^{1/2}X_1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.