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    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!
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    (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)?
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    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
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    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:

    f_X(x) = \displaystyle\int_{-\infty}^{\infty} f_U(u)f_Y(x/u)\dfrac{1}{|u|} du = (\text{constants}) \displaystyle\int_{0.1}^{1.5} \dfrac{1}{u}\exp \left[-\frac{1}{2}\left(\dfrac{x/u-\mu}{\sigma}\right)^2\right] du

    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.
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    (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:

    f_X(x) = \displaystyle\int_{-\infty}^{\infty} f_U(u)f_Y(x/u)\dfrac{1}{|u|} du = (\text{constants}) \displaystyle\int_{0.1}^{1.5} \dfrac{1}{u}\exp \left[-\frac{1}{2}\left(\dfrac{x/u-\mu}{\sigma}\right)^2\right] du

    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!
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    (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.
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    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.
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    (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 p(x) = \int p(x|\mu) p(\mu) d \mu, 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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