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Math Help - Linked probabilities

  1. #1
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    Linked probabilities

    Hi everyone,

    My question seems to be easy, but I can't solve it...
    Here are three random variables :
    - X ~ N(0,sigma_x) : normal distribtution with mean = 0 and variance = sigma_x▓
    - Y ~ N(0,sigma_y)
    - C ~ N(0,1)

    How can the following probability be calculated :
    Pr { X+C<a et Y+C<b } ??? (with a and b two real numbers)
    The aim is to find a result depending on a, b, sigma_x and sigma_y.

    Thanks in advance,
    Have a nice day,
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  2. #2
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    Tiens, t'es Franšais ?

    Are the three variables independent ? (looks like it is, but I prefer to be sure)


    If it is, then you know the joint probability density function of (X,Y,C) (product of the density functions)

    Now find the region associated to \{(x,y,c)\in\mathbb{R}^3 ~:~ x+c<a ~,~ y+c<b\}=D. By "find", I mean get the boundaries for x,y,c so that you can write the triple integral : \iiint_D f_X(x)f_Y(y)f_C(c) ~dx~dy~dc
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  3. #3
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    Linked probabilites...

    Thanks for your reply.

    The idea is to get a result without integrals.
    Because, after, I will have to find a and b such as :
    Pr { X+C<a et Y+C<b } = 0.9

    I tried hard to calculate the integral that you defined in your post, but without any success.
    Do you have any idea?

    P.S.: oui, je suis franšais
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  4. #4
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    Quote Originally Posted by guigui1024 View Post
    Thanks for your reply.

    The idea is to get a result without integrals.
    Because, after, I will have to find a and b such as :
    Pr { X+C<a et Y+C<b } = 0.9

    I tried hard to calculate the integral that you defined in your post, but without any success.
    Do you have any idea?

    P.S.: oui, je suis franšais
    It is not possible to get an exact value...
    You'll have to make approximations, or use a z-table : http://www.science.mcmaster.ca/psych...e/z-table2.jpg

    I'm a bit busy to make the computations, but even so, I think it would be impossible to find a and b. Or at least you may be able to find a constraint...

    Maybe you can make the transformation (x,y,c) -> (u,v,w)=(x+c,y+c,c) to see more clearly the boundaries.

    I don't know :s
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  5. #5
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    Linked probabilites...

    Hello,

    Thanks for your reply.

    Trying to make the transformation you mentionned, I find a ugly piece of calculation which is a kind of :

    <br />
\int^{c=\infty}_{c=-\infty}\int^{x=a-c}_{x=-\infty}\int^{y=b-c}_{y=-\infty}\exp(-(x^2+y^2+c^2))dxdydc<br />

    I can't solve it... I sent a post in the calculation forum.
    Do you have any idea? I mean, without going throug integration, or to calculate easily this integrate...?

    Have a nice day,

    P.S.: yes, I know the z-table, but it seems that I can't use it directly in my case
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  6. #6
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    [1] From the definition of X,Y,C, it seems they are independant.

    [2]Theorem:

    The distribution of muti-dimensional random variable \left(X_1,X_2,\ldots,X_n\right) is norm distribution ,if and only if

    any linear combination of X_1,X_2,\ldots X_n

    L_1X_1+L_2X_2+\ldots L_nX_n

    is one-dimensional norm distribution.

    [3]According to theorem, it's easy to verify the \left(X+C,Y+C\right) is the norm distribution.

    [4]The coefficients between the two random variables cov\left(X+C,Y+C\right) is easy to handle.

    [5]Based on the above statements, it seem easy to deal with a, b
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  7. #7
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    Thank you Min.Lu, your answer is very helpful.
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