Calculating Probability with Random Variables and Constants

[EngWiPy]

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

 

[mfb]

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

 

[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.

 

[EngWiPy]

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.

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

 

[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.

 

[mfb]

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.