## An easy algebra exercise

This is another algebra problem that I came across in one of my statistics blogs.

Show that:

$\displaystyle \frac{\beta (\beta+1)}{(\alpha+\beta) (\alpha+\beta+1)}+\frac{2 \alpha \beta}{(\alpha+\beta) (\alpha+\beta+1)}+\frac{\alpha (\alpha+1)}{(\alpha+\beta) (\alpha+\beta+1)}=1$

The three fractions on the left hand side are from a probability distribution. Hence the sum is 1. The algebra is not bad and is not difficult to verify.

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## A calculus exercise from a probability distribution

Several calculus exercises that are derived from a probability problem are presented here.

Let $\alpha>0$ be a constant.

Evaluate $\displaystyle \int_0^1 (1-e^{-\alpha z}) \ dz + \int_1^{\infty}(e^\alpha-1) e^{-\alpha z} \ dz$.

Evaluate $\displaystyle \int_0^1 z \ (1-e^{-\alpha z}) \ dz + \int_1^{\infty} z \ (e^\alpha-1) e^{-\alpha z} \ dz$.

Evaluate $\displaystyle \int_0^1 z^2 \ (1-e^{-\alpha z}) \ dz + \int_1^{\infty} z^2 \ (e^\alpha-1) e^{-\alpha z} \ dz$.

Background
For anyone who is interested in the origin of these integrals, the first exercise is to verify that the following function is a probability density function (pdf). The second and the third exercises are to find the first and second moment of this probability distribution.

$\displaystyle f_Z(z)=\left\{\begin{matrix} \displaystyle 1-e^{-\alpha z}&\ \ \ \ \ \ 0 \le z <1 \\{\text{ }}& \\{\text{ }}& \\{\displaystyle e^{-\alpha z}(e^\alpha-1)}&\ \ \ \ \ \ 1 \le z <\infty \end{matrix}\right.$

The above pdf $f_Z(z)$ is the independent sum of an exponential distribution and the uniform distribution $U(0,1)$. So the answers of the integrals can be derived from knowing these probability distributions. Otherwise, they are excellent exercises in calculus.

$\displaystyle \int_0^1 (1-e^{-\alpha z}) \ dz + \int_1^{\infty}(e^\alpha-1) e^{-\alpha z} \ dz=1$.

$\displaystyle \int_0^1 z \ (1-e^{-\alpha z}) \ dz + \int_1^{\infty} z \ (e^\alpha-1) e^{-\alpha z} \ dz=\frac{1}{2}+\frac{1}{\alpha}$.

$\displaystyle \int_0^1 z^2 \ (1-e^{-\alpha z}) \ dz + \int_1^{\infty} z^2 \ (e^\alpha-1) e^{-\alpha z} \ dz=\frac{1}{3}+\frac{1}{\alpha}+\frac{2}{\alpha^2}$.

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## An algebra exercise

As I worked on a problem involving the exponential distribution, I had to carry out an algebra derivation, which is a nice stand alone algebra problem. Here it is:

$\displaystyle \frac{\beta}{\beta - \alpha} \ \frac{\alpha}{\alpha-\gamma}+\frac{\alpha}{\alpha-\beta} \ \frac{\beta}{\beta-\gamma}=\frac{\alpha}{\alpha-\gamma} \ \frac{\beta}{\beta-\gamma}$

The exercise is to show that the left hand side equals to the right hand side. For anyone taking algebra or preparing for SAT or ACT, this would be an excellent exercise. You may want to try it on your own first. You can find the derivation beow.

\displaystyle \begin{aligned}\frac{\beta}{\beta - \alpha} \ \frac{\alpha}{\alpha-\gamma}+\frac{\alpha}{\alpha-\beta} \ \frac{\beta}{\beta-\gamma}&=\frac{-\beta}{\alpha-\beta} \ \frac{\alpha}{\alpha-\gamma}+\frac{\alpha}{\alpha-\beta} \ \frac{\beta}{\beta-\gamma} \\& \text{ } \\&=\frac{-\beta}{\alpha-\beta} \ \frac{\alpha}{\alpha-\gamma} \ \frac{\beta-\gamma}{\beta-\gamma}+\frac{\alpha}{\alpha-\beta} \ \frac{\beta}{\beta-\gamma} \ \frac{\alpha-\gamma}{\alpha-\gamma} \\& \text{ } \\&=\frac{-\alpha \ \beta \ (\beta-\gamma)+\alpha \ \beta \ (\alpha-\gamma)}{(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)} \\& \text{ } \\&=\frac{\alpha \ \beta \ (\gamma-\beta+\alpha - \gamma)}{(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)} \\& \text{ } \\&=\frac{\alpha \ \beta \ (\alpha-\beta)}{(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)} \\& \text{ } \\&=\frac{\alpha \ \beta}{(\alpha-\gamma)(\beta-\gamma)} \\& \text{ } \\&=\frac{\alpha}{\alpha-\gamma} \ \frac{\beta}{\beta-\gamma} \end{aligned}

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