Continuous Probability Distribution

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

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

(5.54)

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

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

(5.55)

the mean value of is

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

(5.56)

and the variance of is again

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

(5.57)

If is some function of then

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

(5.58)

Moreover, if and are independent functions of then

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

(5.59)

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

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

(5.60)