Go back
How to calculate radiance emitted?

How to calculate radiance emitted?

Science

Vote Up
Vote Down

Originally posted by Palynka
Ok, so calculate the following probabilities:

P(X = y1)
P(X = y2)
...
P(X = y20)
You give me the answers. (*yawn*)

Vote Up
Vote Down

Originally posted by FabianFnas
You give me the answers. (*yawn*)
No, no, I'm asking you.

1 edit
Vote Up
Vote Down

Originally posted by Palynka
No, no, I'm asking you.
In the continous case every integration beween any real number and the very same number is of course zero. So...?

I have limited time at my disposal, so if you carry on your reasoning instead of waiting for my response every time, then you may come to your conclusion before my time is up.

Vote Up
Vote Down

Originally posted by FabianFnas
I am, right? Well, I'm not a giant.
The problem is that you don't want to learn, not that you're not a giant

1 edit
Vote Up
Vote Down

Originally posted by FabianFnas
In the continous case every integreation beween any real number and the very same number is of course zero. So...?

I have limited time at my disposal, so if you carry on your reasoning instead of waiting for my response every time, then you may come to your conclusion before my time is up.
That's exactly right. The probability that y1 or y2 or any other were drawn was exactly 0, yet they were eventually drawn.

So that an event has probability zero doesn't mean that it cannot happen. Only that it almost surely won't happen. Because that's the beauty of continuous probability distributions (without mass points). Every element has a probability zero of being drawn, so if that means it is impossible for them to be drawn, then no element could ever be drawn.

Get it? Or do you need me to dumb it down a bit more?

Vote Up
Vote Down

Originally posted by KazetNagorra
I do know how to calculate the probability of a continuous variable. You take a probability density function and integrate.
So how come you don't know how to interpret a 0 probability when the variable you're considering is continuous?

Vote Up
Vote Down

Originally posted by adam warlock
So how come you don't know how to interpret a 0 probability when the variable you're considering is continuous?
Come on adam, Kazet has already conceded that point. Fabian, however, let's see how long it takes him.

I'm betting he'll NEVER concede his point. That's why we love him.

Vote Up
Vote Down

Originally posted by adam warlock
So how come you don't know how to interpret a 0 probability when the variable you're considering is continuous?
I do, I worded it badly, you are right. Do you want a cookie?

Vote Up
Vote Down

Originally posted by Palynka
That's exactly right. The probability that y1 or y2 or any other were drawn was exactly 0, yet they were eventually drawn.

So that an event has probability zero doesn't mean that it cannot happen. Only that it almost surely won't happen. Because that's the beauty of continuous probability distributions (without mass points). Every element has a probabilit ...[text shortened]... hen no element could ever be drawn.

Get it? Or do you need me to dumb it down a bit more?
You haven't convinced me yet.

You give me a buch of bas facts that every teenager have a grasp upon. Then you return to your false statement that everything with a probability zero can happen. I say no.
There is a leap between the two statements. In this statement it's a gap that you have to prove to me first.

You say that probability zero can happen. Give me an real world example. Can you find a man with exactly 180 cm (exactly) of length? This is a trick question. Do you see why? 😉
You say probability is zero, nad the sole reason is that there isn't any. It cannot be any, because every man has a length that is more or less near 180 cm, but never exactly at. (Don't forget the trick.)

1 edit
Vote Up
Vote Down

Originally posted by FabianFnas
You haven't convinced me yet.

You give me a buch of bas facts that every teenager have a grasp upon. Then you return to your false statement that everything with a probability zero can happen. I say no.
There is a leap between the two statements. In this statement it's a gap that you have to prove to me first.

You say that probability zero can happ s a length that is more or less near 180 cm, but never exactly at. (Don't forget the trick.)
There are 6 billion people in the world. Every single one of them has a given height. A height, that with the exact same reasoning that you do for 180cm, they would have a probability zero of having.

So, I just gave you 6 billion examples.

Vote Up
Vote Down

Originally posted by KazetNagorra
I do, I worded it badly, you are right. Do you want a cookie?
Originally posted by KazetNagorra
Consider some finite time interval. In this finite time interval, a black body will emit a finite amount of photons. Consider one emitted photon, which has some random wavelength. In the decimal representation of the wavelength, the probability of one decimal corresponding is 10%, at least in the limit of a large numer of decimals behind the comma (so that the actual physics is irrelevant). So the probability of all decimals corresponding to some finite value is simply the limit of 0.1^x as x goes to infinity, in other words, zero. So zero photons of wavelength 788 nm (or whatever) are emitted in this finite time interval. Now take the limit as the time goes to infinity - still zero. It will never happen.

Yes, it is pretty clear that it was just a case of bad wording... 🙄

Vote Up
Vote Down

Originally posted by FabianFnas
Then you return to your false statement that everything with a probability zero can happen.
No one said anything remotely like that in this thread.

Vote Up
Vote Down

Originally posted by adam warlock
No one said anything remotely like that in this thread.
Well spotted.

I never said that, Fabian. In fact, I explicitly said that all events that are impossible have probability zero, but that the converse wasn't true.

Vote Up
Vote Down

Originally posted by Palynka
Well spotted.

I never said that, Fabian. In fact, I explicitly said that all events that are impossible have probability zero, but that the converse wasn't true.
This was said: "events of probability zero may happen".

Perhaps here we have misunderstood eachother.
I say still no. What do you say?

2 edits
Vote Up
Vote Down

Originally posted by FabianFnas
This was said: "events of probability zero may happen".

Perhaps here we have misunderstood eachother.
I say still no. What do you say?
Yes!

Somebody will be born tomorrow. After 25 years, this person will have a certain height x. But, using the same reasoning you did, P(H=x) = 0, because that is true for all x in the support. So an event of probability zero will happen.