Let X~N(2,2) i.e.X is a Gaussian variable with mean=2 and variance=2. Let x>0.
Write P(X>= -x) in terms of the cdf ∅ of the standard Gaussian variable with a positive argument. In other words, your answer be in terms of ∅(g(x)) where g(x) is a function of x which takes only positive values for x>0.
To find the probability P(X ≥ -x), we can express it in terms of the cumulative distribution function (CDF) of the standard Gaussian variable, denoted as Φ(z), where z is a standardized value.
First, let's standardize the variable X using the formula:
Z = (X - μ) / σ
where Z is the standardized variable, X is the original variable, μ is the mean, and σ is the standard deviation.
In our case, X ~ N(2, 2), so μ = 2 and σ = √2. Plugging these values into the formula, we have:
Z = (X - 2) / √2
Next, we need to reframe the expression P(X ≥ -x) in terms of Φ(z). Let's substitute -x into the standardized equation:
Z = (-x - 2) / √2
To reframe it in terms of Φ(z), we need to express z in terms of x. Solving the standardized equation for x, we get:
x = -√2z - 2
Now, substitute this value of x into P(X ≥ -x):
P(X ≥ -x) = P(X ≥ -(-√2z - 2))
Simplifying further, we have:
P(X ≥ -x) = P(X ≥ √2z + 2)
Since X is a Gaussian variable, its CDF can be expressed in terms of Φ(z) as follows:
P(X ≥ √2z + 2) = 1 - Φ(√2z + 2)
Therefore, the probability P(X ≥ -x) in terms of the CDF of the standard Gaussian variable Φ(z) is given by:
P(X ≥ -x) = 1 - Φ(√2z + 2)
Note that the function g(x) mentioned in the question is defined as g(x) = √2z + 2, where z is a standardized value obtained from the original variable X. Since x > 0, we can express √2z + 2 as g(x) = √2z + 2 > 0.
So the final answer is:
P(X ≥ -x) = 1 - Φ(g(x))
where g(x) = √2z + 2, and z is obtained by standardizing X as Z = (X - μ) / σ.