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_x²
- 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.

Have a nice day,

2. Tiens, t'es Français ?

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

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 français

4. Originally Posted by guigui1024

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 français
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

Hello,

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

$\displaystyle \int^{c=\infty}_{c=-\infty}\int^{x=a-c}_{x=-\infty}\int^{y=b-c}_{y=-\infty}\exp(-(x^2+y^2+c^2))dxdydc$

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

6. [1] From the definition of X,Y,C, it seems they are independant.

[2]Theorem:

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

any linear combination of $\displaystyle X_1,X_2,\ldots X_n$

$\displaystyle L_1X_1+L_2X_2+\ldots L_nX_n$

is one-dimensional norm distribution.

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

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

[5]Based on the above statements, it seem easy to deal with a, b