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Problem

I have to prove the following:

Let $X$ and $Y$ be independent continuous random variables with density function $f:\mathbb R\to\mathbb R$. \ Prove that $(X,X+Y)$ is a continuous bivariate random variable with density function $f_{X,X+Y}(x,z)=f(x)f(z-x)$.

My thoughts

Look at $$\mathbb{P}(X\leqslant x,X+Y\leqslant z)=\mathbb{P}(X\leqslant x,Y\leqslant z-x)=\int_{-\infty}^{z-x}\int_{-\infty}^x f_{X,Y}(u,v)\,du\,dv.$$ Since $X$ en $Y$ are independent and have the same density function, we have $f_{X,Y}(u,v)=f(u)f(v)$, so $$=\int_{-\infty}^{z-x}\int_{-\infty}^x f(u)f(v)\,du\,dv=\int_{-\infty}^{z-x} f(v)\left(\int_{-\infty}^x f(u)\,du\right)dv=\int_{-\infty}^{z-x} f(v)\,dv\cdot \int_{-\infty}^x f(u)\,du$$ Now I feel like I am very close, but we can't use the FTC since $f$ is not necessarily continuous (only right continuous). This might be a stupid question, but I don't know how to proceed

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Dr. Heinz Doofenshmirtz
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2 Answers2

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$$ \Pr(X\le x\ \&\ X+Y\le z) = \int_{-\infty}^x \int_{-\infty}^{z\,-\,u} f_{X,\,Y} (u,v) \, dv \, du = \int_{-\infty}^x \int_{-\infty}^{z-u}f(u)f(v) \, dv\,du $$

For any particular value $u$ that the random variable $X$ can assume between $-\infty$ and the value $x$ being considered, the value of $X+Y = u+Y$ is $\le z$ precisely if $Y\le z-u.$

Next we have \begin{align} & \int_{-\infty}^x \int_{-\infty}^{z\,-\,u} f(u) f(v) \, dv \, du = \int_{-\infty}^x \int_{-\infty}^z f(u)f(v-u) \, dv\,du \\[10pt] \text{because } & \int_{-\infty}^{z-u} f(v)\,dv = \int_{-\infty}^z f(v-u)\, du \text{ by routine substitution.} \end{align} Then we will have $$ \Pr(X\le x\ \&\ X+Y\le z) = \int_{-\infty}^x \int_{-\infty}^z f(u) f(v-u) \,dv\,du $$ and that implies that the function being integrated on the right side is the density of $(X,\,X+Y).$

Michael Hardy
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  • But what about the false start comment by drhab? I don't understand it anymore, could you elaborate by posting a proof ? – Dr. Heinz Doofenshmirtz Jan 18 '18 at 21:50
  • @HeinzDoofenschmirtz : I've revised this. $\qquad$ – Michael Hardy Jan 20 '18 at 02:22
  • The 'routine substitution' is the part that boggles me. Namely, you do the substitution $u=v+u$ on the integral $\int_{-\infty}^{z-u} f(v),dv$, which doesn't make sense? Could you explain further? – Dr. Heinz Doofenshmirtz Jan 20 '18 at 09:55
  • @HeinzDoofenschmirtz : You have $\displaystyle \int_{-\infty}^{z-u} f(v),dv. $ Now let $w=v+u,$ so that $v=w-u$ and $\dfrac{dw}{dv} = 1,$ and so $dw = dv.$ As $v$ goes from $-\infty$ to $z-u,$ then $w$ goes from $-\infty$ to $z.$ So the integral becomes $\displaystyle \int_{-\infty}^z f(w-u) , dw.$ And since $w$ is a bound variable, it can be freely renamed, so that for example, this last integral could be called $\displaystyle \int_{-\infty}^z f(r-u), dr$ or $\displaystyle \int_{-\infty}^z f(s-u) , ds.$ Thus it is $\displaystyle \int_{-\infty}^z f(v-u) , dv. \qquad$ – Michael Hardy Jan 21 '18 at 14:28
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The continuity of $X+Y$ is obvious from that of $X$ and $Y$. Let $g:\Bbb{R}^2\to\Bbb{R}$ be a bounded continuous function.

In the second step, make a change of variables $z = x+y$. Thanks to Fubini's theorem, changing order of integration is allowed.

\begin{align} \Bbb{E}[g(X,X+Y)] &= \int_{\Bbb{R}^2} g(x,x+y) f_X(x) f_Y(y) \,dx\,dy \\ &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} g(x,x+y) f_Y(y) \,dy \right) f_X(x) \,dx \\ &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} g(x,z) f_Y(z-x) \,dz \right) f_X(x) \,dx \\ &= \int_{\Bbb{R}^2} g(x,z) f_X(x) f_Y(z-x) \,dx \,dz \end{align}

Therefore, $(X,X+Y)$ has density $f_{X,X+Y}(x,z) = f_X(x) f_Y(z-x)$.

In particular, when $f_X = f_Y = f$, $f_{X,X+Y}(x,z) = f(x) f(z-x)$.

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