Continuous Probability Distribution

Consider some physical system $A$. Suppose that a measurement of a particular physical property of this system, $x$, can result in a continuous range of different outcomes such that $-\infty<x<\infty$. Now, we would expect the probability that a measurement of $x$ yields a result in the range $x$ to $x+dx$ to be proportional to $dx$, in the limit that $dx\rightarrow 0$. (See Section 5.1.7.) Hence, we can define the probability density, $P(x)$, such that the probability of a measurement of $x$ yielding a result in the range $x$ to $x+dx$ is $P(x)\,dx$. A simple extension of the result (5.3) yields the normalization condition,

$$\int_{-\infty}^\infty P(x)\,dx = 1.$$ (5.54)

It follows, from a straightforward extension of the results in Section 5.1.3 that the mean value of $x$ is

$$\langle x\rangle = \int_{-\infty}^\infty P(x)\,x\,dx,$$ (5.55)

the mean value of $x^2$ is

$$\left\langle x^2\right\rangle = \int_{-\infty}^\infty P(x)\,x^2\,dx,$$ (5.56)

and the variance of $x$ is again

$$\left\langle({\mit\Delta} x)^2\right\rangle= \left\langle x^2\right\rangle -\langle x\rangle^2.$$ (5.57)

If $X(x)$ is some function of $x$ then

$$\left\langle X\right\rangle = \int_{-\infty}^\infty P(x)\,X(x)\,dx.$$ (5.58)

Moreover, if $X(x)$ and $Y(x)$ are independent functions of $x$ then

$$\left\langle X+ Y\right\rangle = \int_{-\infty}^\infty P(x)\left[... ...\infty P(x)\,Y(x)\,dx= \left\langle X\right\rangle+\left\langle Y\right\rangle.$$ (5.59)

Finally, in some situations it is convenient to use a probability density, $P(x)$, that does not satisfy the normalization condition (5.54). In such situations,

$$\left\langle X\right\rangle =\frac{ \int_{-\infty}^\infty P(x)\,X(x)\,dx}{ \int_{-\infty}^\infty P(x)\,dx}.$$ (5.60)