# Thread: Cumulative distribution function question, involves calculus

1. ## Cumulative distribution function question, involves calculus

I'm trying to teach myself mathematical statistics to help ground my more applied statistical work in the social sciences. I've been working through some problems in an old textbook that has no known solutions manual and have been doing pretty well, but sometimes get stuck on the calculus. So, this problem could just as easily go in the calculus thread, at least if my understanding of it is correct.

Problem:
Suppose that the c.d.f. $F_Xx$ can be written as a function of $(x-\alpha)/\beta$ where $\alpha$ and $\beta>0$ are constants; that is, $x, \alpha,$ and $\beta$ appear in $F_X(\cdot)$ only in the indicated form.

Prove that if $\alpha$ is increased by $\Delta \alpha$, then so is the mean of $X$.

Solution:
I know that the mean of X is $\int_{-\infty}^\infty xf_X(x)dx$ where $f_X(\cdot)$ is the p.d.f. and the derivative of the c.d.f. is the p.d.f., but when I substitute the derivative of the c.d.f. into the definition of mean I get an integral I can't solve (because there's not enough information about the actual function or because my calculus is so rusty, probably the latter).

Thanks in advance for shedding some light on this for me!

BC

2. Originally Posted by BCarroll05
I'm trying to teach myself mathematical statistics to help ground my more applied statistical work in the social sciences. I've been working through some problems in an old textbook that has no known solutions manual and have been doing pretty well, but sometimes get stuck on the calculus. So, this problem could just as easily go in the calculus thread, at least if my understanding of it is correct.

Problem:
Suppose that the c.d.f. $F_Xx$ can be written as a function of $(x-\alpha)/\beta$ where $\alpha$ and $\beta>0$ are constants; that is, $x, \alpha,$ and $\beta$ appear in $F_X(\cdot)$ only in the indicated form.

Prove that if $\alpha$ is increased by $\Delta \alpha$, then so is the mean of $X$.

Solution:
I know that the mean of X is $\int_{-\infty}^\infty xf_X(x)dx$ where $f_X(\cdot)$ is the p.d.f. and the derivative of the c.d.f. is the p.d.f., but when I substitute the derivative of the c.d.f. into the definition of mean I get an integral I can't solve (because there's not enough information about the actual function or because my calculus is so rusty, probably the latter).

Thanks in advance for shedding some light on this for me!

BC
By definition, $0 \leq F(x) \leq 1$. Now note that [tex]F(\alpha) = 0 and F(\alpha + \beta) = 1[tex].

It follows that $f(x) = \frac{dF}{dx} = \frac{1}{\beta}$ for $\alpha \leq x \leq \alpha + \beta$ and zero elsewhere.

It follows that the mean is $\alpha + \frac{\beta}{2}$ and the answer to the question trivially follows from this.