Finding the distribution of a linear combination of r.v.'s.

[Eclair_de_XII]

Homework Statement

"Let $X,Y$ be independent r.v.'s (EDITED) normally distributed with $\mu=0,\sigma^2=1$. Find the distribution of $W=2X-Y$.

Homework Equations

"If $X,Y$ are independent, then if $Z=X+Y$, $f_{Z}=\int_{\mathbb{R}} f_X(x)f_Y(z-x)\, dx$.

The Attempt at a Solution

First, what I did was find the distribution of $U=2X$ and $V=-Y$.

$P(U\leq u)=P(2X\leq u)=P(X\leq \frac{u}{2})=\int_{-\infty}^{\frac{u}{2}} f_X(x)\, dx$. Let $s=2x$, and $\frac{ds}{2}=dx$. Then $\frac{u}{2}\mapsto u$, and $-\infty \mapsto -\infty$. So $P(X\leq \frac{u}{2})=\frac{1}{2} \int_{-\infty}^u f_X(s)\, ds$ and $f_U(u)=\frac{1}{2}f_X(u)$. Similarly, $f_V(v)=-f_Y(v)$. I know that pdf's are always non-negative, which is partially why I got stuck.

Next, I let $W=U+V$ so that $f_W(w)=\int_{\mathbb{R}} f_U(u)f_V(w-u)\, du$, and this is where I got stuck, next. I feel like I'm supposed to substitute $u$ for $x$ here, but I can't remember how to do this...

 

[Ray Vickson]

Homework Statement

"Let $X,Y$ be normally distributed with $\mu=0,\sigma^2=1$. Find the distribution of $W=2X-Y$.

Homework Equations

"If $X,Y$ are independent, then if $Z=X+Y$, $f_{Z}=\int_{\mathbb{R}} f_X(x)f_Y(z-x)\, dx$.

The Attempt at a Solution

First, what I did was find the distribution of $U=2X$ and $V=-Y$.

$P(U\leq u)=P(2X\leq u)=P(X\leq \frac{u}{2})=\int_{-\infty}^{\frac{u}{2}} f_X(x)\, dx$. Let $s=2x$, and $\frac{ds}{2}=dx$. Then $\frac{u}{2}\mapsto u$, and $-\infty \mapsto -\infty$. So $P(X\leq \frac{u}{2})=\frac{1}{2} \int_{-\infty}^u f_X(s)\, ds$ and $f_U(u)=\frac{1}{2}f_X(x)$. Similarly, $f_V(v)=-f_Y(y)$. I know that pdf's are always non-negative, which is partially why I got stuck.

Next, I let $W=U+V$ so that $f_W(w)=\int_{\mathbb{R}} f_U(u)f_V(w-u)\, du$, and this is where I got stuck, next. I feel like I'm supposed to substitute $u$ for $x$ here, but I can't remember how to do this...

The given information is not sufficient to determine an explicit answer. We also need to know the covariance between $X$ and $Y$, or whether $X,Y$ are independent.

 

[Eclair_de_XII]

Oh, right. I forgot to mention that they're independent.

 

[Eclair_de_XII]

Anyway, this is what I have, in addition to the work done in the opening post...

$f_W(w)=\int_{\mathbb{R}} f_U(u)f_V(w-u)\, du=-\frac{1}{2} \int_{\mathbb{R}} f_X(u)f_Y(w-u)\, du=-\frac{1}{4\pi} \int_{\mathbb{R}} e^{-\frac{1}{2}u^2}e^{-\frac{1}{2}(w-u)^2}\, du\\ =-\frac{1}{4\pi} \int_{\mathbb{R}} e^{-\frac{1}{2}(2u^2-2wu+w^2)}\, du=-\frac{1}{4\pi} e^{-\frac{1}{2}w^2} \int_{\mathbb{R}} e^{-(u^2-wu)}\\ =-\frac{1}{4\pi} e^{-\frac{1}{2}w^2} \int_{\mathbb{R}} e^{-[(u-\frac{1}{2}w)^2-\frac{1}{4}w^2]}=-\frac{1}{4\pi} e^{-\frac{1}{4}w^2} \int_{\mathbb{R}} e^{-(u-\frac{1}{2}w)^2}=-\frac{1}{4\pi} e^{-\frac{1}{4}w^2}$, which isn't a pdf, since it's always negative.

 

[member 587159]

Do you already know a formula for linear combinations of independent distributions?

I suggest to abstract the problem, solve this one and come back to your original problem.

 

[Eclair_de_XII]

Do you already know a formula for linear combinations of independent distributions?

I do not.

 

[Ray Vickson]

Anyway, this is what I have, in addition to the work done in the opening post...

$f_W(w)=\int_{\mathbb{R}} f_U(u)f_V(w-u)\, du=-\frac{1}{2} \int_{\mathbb{R}} f_X(u)f_Y(w-u)\, du=-\frac{1}{4\pi} \int_{\mathbb{R}} e^{-\frac{1}{2}u^2}e^{-\frac{1}{2}(w-u)^2}\, du\\ =-\frac{1}{4\pi} \int_{\mathbb{R}} e^{-\frac{1}{2}(2u^2-2wu+w^2)}\, du=-\frac{1}{4\pi} e^{-\frac{1}{2}w^2} \int_{\mathbb{R}} e^{-(u^2-wu)}\\ =-\frac{1}{4\pi} e^{-\frac{1}{2}w^2} \int_{\mathbb{R}} e^{-[(u-\frac{1}{2}w)^2-\frac{1}{4}w^2]}=-\frac{1}{4\pi} e^{-\frac{1}{4}w^2} \int_{\mathbb{R}} e^{-(u-\frac{1}{2}w)^2}=-\frac{1}{4\pi} e^{-\frac{1}{4}w^2}$, which isn't a pdf, since it's always negative.

Why do you persist in making obvious errors? Start from the basics, every time, and you will never go wrong! If $V = -Y$, then for small $h > 0$ we have $P(v < V < v + h) = P(v < -Y < v + h) = P(-v-h < Y < -v) = h f_Y(-v),$ because we are looking at an interval of length $h$ going from $-y-h$ up to $-y$. We do NOT get $-h f(-v)$ because the length of the interval is positive, not negative.

You really need to guard against letting your own notation fool you, and you do that by thinking, not just substituting into formulas too quickly.