# Dirac delta function with a sum as the argument

I'm reading "First steps in random walks" by Klafter and Sokolov, and I don't understand this step involving the Dirac delta function. They want to obtain the probability density of having a walker at $x$ after $n$ steps: $P_n(x)$. They begin with

$$P_n(x)=\int_{-\infty}^\infty dy P_{n-1}(y)p(x-y), \qquad (1)$$ where $p(x)$ is the probability density to take a step of length $x$. Then they reiterate (1) and obtain (making a change of variables, I suppose)

$$P_n(x)=\int_{-\infty}^\infty\cdots \int_{-\infty}^\infty dx_1\cdots dx_{n-1} p(x_1)p(x_2)\cdots p(x-x_{n-1}), \qquad (2)$$

and then they introduce the Dirac delta function as a "formal trick", they say:

$$P_n(x)=\int_{-\infty}^\infty\cdots \int_{-\infty}^\infty dx_1\cdots dx_n p(x_1)p(x_2)\cdots p(x_n) \delta \left(\sum_{i=1}^nx_i-x \right), \qquad (3)$$

My question is: how can we show (heuristically) the equivalence between (2) and (3)? I'm used to Dirac delta functions of the form $\delta(x-x_0)$, but not of the form at (3).

EDITED

Maybe it would help to notice that $P_n(x)$ is the probability density of the random variable $X_n:=\sum_{i=1}^nx_i$.

• Shouldn't the $n$th term in (2) be $p(x-x_{n-1}-\cdots-x_1)$?
– anon
Jun 5 '15 at 18:50
• @anon You say that because it will solve the transition between 2 and 3 or because you made the change of variables to obtain 2 from 1? Jun 5 '15 at 19:23

From the get-go we should be able to tell the equation (2) is incorrect, since it equals

$$\left(\int_{-\infty}^\infty p(x_1)dx_1\right)\cdots\left(\int_{-\infty}^\infty p(x_{n-2})dx_{n-2}\right)\int_{-\infty}^\infty p(x_{n-1})p(x-x_{n-1})dx_{n-1} \tag{2-bad}$$

which is $1\cdots 1\cdot P_2(x)$ instead of the $P_n(x)$ it's supposed to be.

The correct form is something that can be intuited. Since step sizes are independent events, the probability density of having step sizes $(x_1,\cdots,x_n)$ should be $p(x_1)\cdots p(x_{n-1})p(x_n)$ which we should then integrate over the hyperplane defined by the equation $x_1+\cdots+x_n=x$ in order to obtain the probability the $n$ steps sum to $x$. Such a domain of integration can be parametrized by letting the variables $x_1,\cdots,x_{n-1}$ roam freely and then setting $x_n=x-(x_1+\cdots+x_{n-1})$, yielding

$$\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty p(x_1)\cdots p(x_{n-1})p\big(x-(x_1+\cdots+x_{n-1})\big) dx_{n-1}\cdots x_1. \tag{2-good}$$

We can derive this from (1) using substitution if we wish. Observe

$$P_n(x)=\int_{-\infty}^\infty p(x-u_1)P_{n-1}(u_1)du_1=\int_{-\infty}^\infty\int_{-\infty}^\infty p(x-u_1)p(u_1-u_2)P_{n-2}(u_2)du_2du_1$$

$$\cdots =\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty p(x-u_1)p(u_1-u_2)\cdots p(u_{n-2}-u_{n-1})P_1(u_{n-1})du_{n-1}\cdots du_1$$

and of course $P_1=p$. Make a change of variables

$$\begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_{n-2} \\ x_{n-1}\end{bmatrix}=\begin{bmatrix}u_1-u_2 \\ u_2-u_3 \\ \vdots \\ u_{n-2}-u_{n-1} \\ u_{n-1}\end{bmatrix}=\begin{bmatrix} 1 & -1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & -1 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & -1 \\ 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix} \begin{bmatrix}u_1 \\ u_2 \\ \vdots \\ u_{n-2} \\ u_{n-1}\end{bmatrix}.$$

The Jacobian determinant of this upper triangular matrix is $1$, and $u_1=x_1+\cdots+x_{n-1}$, so the integral becomes precisely the one written in $(\text{2-good})$.

• This answer is "2-good". BTW do you know any book containing all the tricks involving the Dirac delta function? @anon Jun 8 '15 at 22:33
• @Anuar Sorry, I do not. I can't think of a trick I would use that's not on Wikipedia, though.
– anon
Jun 9 '15 at 2:02

Equation (3) is a generalized method of constructing histograms, a.k.a. probability density functions (pdf).

Say a magnetic field intensity varies in the $x$ direction: $B = b\left(x\right)$ and you want to construct a histogram of the field over the 1d line, which equals the probability density function of B when x is chosen randomly with from the distribution $f\left(x\right)$ the pdf of the magnetic field is given by, with the definition that $f\left(x\right)dx$ is the probability of $x$ being between $x$ and $x+dx$ and the probability of witnessing a field intensity between $B$ and $B+dB$ at that location is given by $p\left(B\right)dB$

$p\left ( B \right )=\int dx f\left(x\right) \delta \left ( B-b\left ( x \right )\right )$

so if you imagine a field value of $B^{\star}$ that exceeds/deceeds the maximum/minimum of $b\left(x\right)$, it will never be experienced at any location $x$ since $b\left(x\right)$ will never equal $B^{\star}$ and hence the delta function $\delta\left(B^{\star}-b\left(x\right)\right)$ and thus $p\left(B^{\star}\right)$ will always be zero since the $\delta$-fxn argument will not equal zero. On the other hand, if there is a field value $B_0$ at two points, $x_1$ and $x_2$, the value of $p\left(B_0\right)$ should equal $f\left(x_1\right) + f\left(x_2\right)$ ... and since the delta function will integrate to unity at these locations for that field value, the equation is correctly constructed.

• Sorry but this is hardly mathematics at all (and no, histograms are not "a.k.a. probability density functions").
– Did
Aug 17 '17 at 15:27
• @Did, you're wrong. Aug 17 '17 at 15:37