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# Probability of 2 Statements (no independence) watch

1. Hey guys,

I want to determine the probability that the following statements occur:
Code:
a<X
a>Y-b
a and b are fixed values, X and Y are normal distributed random variables.

Taking both statements solely, it is easy for me to calculate the probability with the cumulative distribution function:
Code:
P(a<X) = 1-P(X<a) = 1-(a-E[X])/Sigma(X)

P(a>Y-b) = P(Y<a+b) = (a+b-E[Y])/Sigma(Y)
But in my case, a is a fraction of b and X is a fraction of Y (again, calculated by a uniform distributed random coefficient between 0.1 and 1.5).

Therefore, both probabilities are not independent and I cannot simply multiply them, right?

Any clues how I can come up with the combined probability P(Y-b<a<X)?

Thanks a lot for your ideas in advance!
2. (Original post by KoJoe)
But in my case, a is a fraction of b and X is a fraction of Y (again, calculated by a uniform distributed random coefficient between 0.1 and 1.5).
Are you saying that a is the same random fraction of b as X is of Y? That is a = u b and X = u Y for u a uniform variate in (0.1, 1.5)?
3. Hi Gregorius,
Sorry for not being specific.
X = u Y [u is a uniform variate in (0.1, 1.5)]
but for a and b it is different.
0<=a<=b, but nothing is known about the distribution
4. I'm not sure if I'm being daft but isn't it the case that X cannot be normally distributed if X=UY with U and Y as defined above? Since the density function will be given by:

Which, even in the case of standard normal, doesn't seem to come out much like the PDF of a normally distributed random variable.

[disclaimer: probability is not my thing so I may have missed the point]

EDIT: That said, if we ignore the claim that X is normally distributed, you can probably still reach an answer if you find the above pdf (which doesn't seem very nice, to be brutally honest), and then the joint distribution of X and Y......-many more dots-. So there's either a much nicer way that I'm missing (likely), or this probability is just downright nasty to calculate.
5. (Original post by Farhan.Hanif93)
I'm not sure if I'm being daft but isn't it the case that X cannot be normally distributed if X=UY with U and Y as defined above? Since the density function will be given by:

Which, even in the case of standard normal, doesn't seem to come out much like the PDF of a normally distributed random variable.

[disclaimer: probability is not my thing so I may have missed the point]
Yes, this is true. I expect that what is meant (is this true, OP?) is that X is normally distributed conditional upon the value of u. It looks to me that the way to attack this problem is to do the calculations conditional upon the value of u and then to integrate u out at the end. May have a go later, but have a busy day at work from now on!

Another question for OP is where has this question come from? Is it an exercise from a course or is it a problem that has come up IRL? If it were the latter, I would be sore tempted to answer it by simulation!
6. (Original post by Gregorius)
Yes, this is true. I expect that what is meant (is this true, OP?) is that X is normally distributed conditional upon the value of u.
Ah, that seems to be far more sensible.
7. Hey guys,
great brainstorming!

In fact, I wanted to keep it simple for everyone. But obviously you are right, my simplicity resulted in a more complex problem.

So X and Y are both normally distributed.
X = u*Y is actually NOT true
E[X] = u*E[Y] and based on those expected values X and Y will be normally distributed.

The question comes from an optimization I want to do, the probability will be in the objective function of a nonlinear problem.
8. (Original post by KoJoe)
Hey guys,
great brainstorming!

In fact, I wanted to keep it simple for everyone. But obviously you are right, my simplicity resulted in a more complex problem.

So X and Y are both normally distributed.
X = u*Y is actually NOT true
E[X] = u*E[Y] and based on those expected values X and Y will be normally distributed.
It looks as if you are saying that (using Bayesian terminology) that you're putting a uniform prior distribution on the mean of a normal distribution. You can then get , but in this case, you'll end up with an expression in terms of the difference of error functions.

The question comes from an optimization I want to do, the probability will be in the objective function of a nonlinear problem.
Might be an idea to post the original problem?

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