# B Finding this probability

1. Jul 26, 2016

### S_David

Hello all,

I have a question:

Suppose I want to find the following probability:

$$Pr\left[\frac{\alpha_1}{\alpha_2+1}\leq\gamma\right]$$

where $\alpha_i$ for i=1, 2 is a random variable, whatever the distribution is, and $\gamma$ is a constant. Can I write it as

$$Pr\left[\frac{\alpha_1}{\alpha_2+1}\leq\gamma\right]=\int_{\alpha_2}\Pr\left[\alpha_1\leq\gamma(\alpha_2+1)\right]\,f_{\alpha_2}(\alpha_2)\,d\alpha_2$$

where $f_{\alpha_2}(\alpha_2)$ is the p.d.f of the random variable $\alpha_2$, or I need to average over the distribution of $\gamma(\alpha_2+1)$?

Thanks

2. Jul 26, 2016

### Staff: Mentor

The integral is exactly this "average". Should work like that, assuming α2+1 > 0.

3. Jul 26, 2016

### mathman

$Pr(\frac{\alpha_1}{\alpha_2 +1}\le \gamma)=Pr(\alpha_1 \le (\alpha_2+1)\gamma)=Pr(\alpha_1-\alpha_2 \times \gamma \le \gamma)$.. The last expression is in the form of a sum of (presumed independent) random variables. There is a standard expression, involving convolution of the individual distributions. This assumes $\alpha_2+1 \ge 0$.

4. Jul 26, 2016

### S_David

Thanks. Could you give more details on how to find your last probability in terms of the individual pdfs of the independent random variables?

5. Jul 27, 2016

### chiro

Hey S_David.

If you have a function of a random variable you can use the transformation theorem to get the new PDF for the new random variable.

For the sums look at convolution or probability generating functions if they are independent or use the joint and/or conditional distributions to get the probability.

6. Jul 27, 2016

### Staff: Mentor

Just to clarify that: the approach in post 1 is fine, the approach with the convolution will lead to the same thing, just expressed in a different way.