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| − | I don't understand how the P[U< | + | I don't understand how the P[U<=F(x)] = F(x)can someone explain this? |
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| + | //Comment | ||
| + | This can be said because it was given to us that the function is a "non-decreasing function" and that U is a uniform random variable. This is enough information to make this step. So, | ||
| + | :<math> | ||
| + | \begin{align} | ||
| + | & {} \Pr(U \leq F(x)) = F(x)\quad \text{(because }\Pr(U \leq y) = y,\text{ since }U\text{ is uniform on the unit interval)} | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Which can be stated more simply that if a probability is increased then the probability that your variable will be less than that probability will be the exact amount that you increased it by; but, only if the probability is always increasing and never decreases. | ||
| + | |||
| + | I apologize this explanation may not be as helpful as you hoped, but it is very hard for me to explain without a picture or in person. | ||
| + | |||
| + | Sincerely, | ||
| + | |||
| + | Jared McNealis | ||
Latest revision as of 04:42, 21 October 2008
I don't understand how the P[U<=F(x)] = F(x)can someone explain this?
//Comment
This can be said because it was given to us that the function is a "non-decreasing function" and that U is a uniform random variable. This is enough information to make this step. So,
- $ \begin{align} & {} \Pr(U \leq F(x)) = F(x)\quad \text{(because }\Pr(U \leq y) = y,\text{ since }U\text{ is uniform on the unit interval)} \end{align} $
Which can be stated more simply that if a probability is increased then the probability that your variable will be less than that probability will be the exact amount that you increased it by; but, only if the probability is always increasing and never decreases.
I apologize this explanation may not be as helpful as you hoped, but it is very hard for me to explain without a picture or in person.
Sincerely,
Jared McNealis
