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Here is the question:

Let $X$ and $Y$ be independent, continuous r.v.s with PDFs $f_X$ and $f_Y$ respectively, and let $T=X+Y$. Find the join PDF of $T$ and $X$, and use this to give a proof that $f_T(t)=\int_{-\infty}^{\infty}f_X(x)f_Y(t-x)dx$.

Here is my attempt:

First find the CDF of $X,T$: \begin{align} F_{X,T}(x,t) &=P(X\leq x,T \leq t)\\ &=P(X\leq x,X+Y \leq t)\\ &=P(X\leq x,Y \leq t-X)\\ &=P(X\leq x,Y \leq t-x)\\ &=P(X\leq x)P(Y \leq t-x)\\ &=F_X(x)F_Y(t-x) \end{align}

Thus, we can get the PDF by taking the derivative with respect to $t$: \begin{align} f_{X,T}(x,t) &=\frac{\partial^2}{\partial x \partial t}F_{X,T}(x,t)\\ &=\frac{\partial}{\partial x}(F_X(x)f_Y(t-x)) \end{align}

To get the PDF of $T$: \begin{align} f_T(t) &=\int_{-\infty}^{\infty}\frac{\partial}{\partial x}(F_X(x)f_Y(t-x))dx\\ &=F_X(x)f_Y(t-x) \end{align}

This clearly is not the convolution, does anyone know why this is wrong?

EDIT

Alright, so turns out I just forgot to evaluate the integral. Then, suppose I did this: $$f_T(t)=\int_{-\infty}^{\infty}\frac{\partial}{\partial x}(F_X(x)f_Y(t-x))dx$$

Apply chain rule and we get: $$\int_{-\infty}^{\infty}(f_X(x)f_Y(t-x)+F_X(x)\frac{\partial}{\partial x}f_Y(t-x))dx$$

Now this almost looks like the convolution, is the second term some how zero or have I made a mistake?

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I am not quite sure, but I think your error is $P(X \le x, Y \le t-X) = P(X \le x, Y \le t-x)$. The most you can say here is "$\le$"; I don't think equality holds.


It suffices to show $f_{X,T}(x,t) = f_X(x) f_Y(t-x)$; then, integrating over $x$ gives the desired form of $f_T(t)$.

My computation below is a little unrigorous because I condition on an event $\{X=u\}$ that has zero probability. This issue can be resolved (see here for example), but I am curious if others have a better way of presenting this computation.

\begin{align} P(X \le x, X+Y \le t) &= \int_{-\infty}^\infty P(X \le x, X+Y \le t \mid X=u) f_X(u) \mathop{du}\\ &= \int_{-\infty}^x P(Y \le t-u) f_X(u) \mathop{du}\\ &= \int_{-\infty}^x F_Y(t-u) f_X(u)\\ \frac{\partial^2}{\partial x \partial t}P(X \le x, X+Y \le t) &= \frac{\partial}{\partial t} F_Y(t-x) f_X(u) \mathop{du}\\ &= f_Y(t-x) f_X(u). \end{align}

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  • $\begingroup$ Wait, so $\int_{-\infty}^{\infty}P(X \leq x \mid X=u)f_X(u)=\int_{-\infty}^{x}f_X(u)$? What does $P(X \leq x \mid X=u)$ even mean? $\endgroup$ – Zoom Bee May 31 '16 at 4:04
  • $\begingroup$ One more question is how did you get from line 3 to line 4? How did you get rid of the integral, if you can explain this to me that would be awesome! $\endgroup$ – Zoom Bee May 31 '16 at 4:12
  • $\begingroup$ @ZoomBee The integral disappears by applying the Fundamental Theorem of Calculus. I think $P(X \le x \mid X=u)$ is $1$ if $u \le x$ and zero otherwise. I suggest you check out copper.hat's answer, as it avoids this messy business of conditioning. $\endgroup$ – angryavian May 31 '16 at 4:22
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\begin{eqnarray} P [ X+Y \le \alpha] &=&\int_{x=-\infty}^\infty \int_{y=-\infty}^\infty 1_{\{(x,y) | x+y \le \alpha \}} ((x,y)) f_X(x) f_Y(y)dy dx \\ &=& \int_{x=-\infty}^\infty \int_{y=-\infty}^{\alpha-x} f_X(x) f_Y(y) dy dx \\ &=& \int_{x=-\infty}^\infty \int_{y=-\infty}^{\alpha} f_X(x) f_Y(y-x) dy dx \\ &=& \int_{y=-\infty}^{\alpha} \int_{x=-\infty}^\infty f_X(x) f_Y(y-x) dy dx \\ &=& \int_{y=-\infty}^{\alpha} f_T(y) dy \\ \end{eqnarray} where $f_T(t) = \int_{x=-\infty}^\infty f_X(x) f_Y(t-x) dx$

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the last step is wrong. you are integrating over $x$, so the end result of the (definite) integral does not depend on $x$. looks like you forgot to evaluate it.

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  • $\begingroup$ Hmmm, if that is the case, I decided to apply the chain rule to the inside and it sort of looks like the convolution but not exactly. I updated my question, do you know if the second term is zero or I made a mistake? $\endgroup$ – Zoom Bee May 31 '16 at 2:37
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Following your initial lines, \begin{multline*} F_{X,T}(x,t) = P(X\le x, T\le t) = P(X\le x, X+Y\le t) = P(X\le x, X+Y\le t - X) \\ = \int_{-\infty}^{x} P(X\le x, X+Y\le t - X|X=z) f_X(z) dz = \int_{-\infty}^{x} P(z\le x, Y\le t - z|X=z) f_X(z) dz \\ = \int_{-\infty}^{x} P(Y\le t - z|X=z) f_X(z) dz = \int_{-\infty}^{x} P(Y\le t - z) f_X(z) dz \\ = \int_{-\infty}^{x} F_Y(t-z) f_X(z) dz. \end{multline*}

Now recall that $F_T(t) = \lim_{x\to\infty} F_{X,T}(x,t)$ and apply it on the above: \begin{multline*} F_T(t) = \lim_{x\to\infty} F_{X,T}(x,t) = \lim_{x\to\infty} \int_{-\infty}^{x} F_Y(t-z) f_X(z) dz \\ = \int_{-\infty}^{\infty} F_Y(t-z) f_X(z) dz = \int_{-\infty}^{\infty} F_Y(t-x) f_X(x) dx. \end{multline*}

Now $f_T(t) = F_T'(t)$, so, skipping lazily all due justifications, we have \begin{multline*} f_T(t) = \frac{d}{dt} F_T(t) = \frac{d}{dt} \left(\int_{-\infty}^{\infty} F_Y(t-x) f_X(x) dx\right) = \int_{-\infty}^{\infty} \left(\frac{d}{dt} F_Y(t-x)\right) f_X(x) dx = \int_{-\infty}^{\infty} \left( \frac{\partial}{\partial y} F_Y(t-x) \frac{d}{dt} (t-x) \right) f_X(x) dx = \int_{-\infty}^{\infty} f_Y(t-x) f_X(x) dx, \end{multline*} as intended initially.

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