What is the probability that the sum of a standard normal random variable, X, and another random variable, Y, is less than or equal to zero?
To find the probability that the sum of X and Y is less than or equal to zero, we need to determine the joint distribution of X and Y. Since X is a standard normal random variable, we know that it has a mean of 0 and a standard deviation of 1.
However, we don't have any information about the distribution of Y. If Y is also a standard normal random variable with a mean of 0 and a standard deviation of 1, then we can calculate the probability by using the properties of the normal distribution.
Let Z = X + Y be the sum of X and Y. If both X and Y are standard normal random variables, then Z will also be a normal random variable with mean μZ = μX + μY = 0 + 0 = 0, and standard deviation σZ = √(σX^2 + σY^2) = √(1^2 + 1^2) = √2.
To find the probability P(Z ≤ 0), we can standardize the variable Z by subtracting the mean and dividing by the standard deviation. Let's denote the standardized variable as Z*.
Z* = (Z - μ)/σ = (Z - 0)/√2 = Z/√2
Now we need to find the probability P(Z* ≤ 0). Since Z* follows a standard normal distribution, we can look up this probability in a standard normal distribution table or use statistical software.
Using the standard normal distribution table or software, we find that the probability P(Z* ≤ 0) is 0.5.
Therefore, if both X and Y are standard normal random variables, the probability that the sum of X and Y is less than or equal to zero is 0.5.
To find the probability that the sum of a standard normal random variable X and another random variable Y is less than or equal to zero, we need to consider the distribution of the sum of two random variables.
When two random variables are summed, their resulting distribution is the convolution of their individual distributions. In this case, since X is a standard normal random variable, its distribution is a standard normal distribution with mean 0 and standard deviation 1.
Let's assume that Y is another random variable with its own distribution, denoted as f(y). To find the probability that X + Y ≤ 0, we need to evaluate the convolution of the distributions of X and Y, and then calculate the area under the resulting density function up to and including zero.
Given that X has a standard normal distribution, its probability density function (pdf) is given by:
f(x) = (1 / √(2π)) * e^(-x^2 / 2)
To find the convolution of X and Y, we need to integrate the following expression:
∫[f(x) * f(y-x)] dx
where f(y-x) is the pdf of Y.
Since we do not have information about the specific distribution of Y, it is not possible to provide an exact solution to this problem. The answer will depend on the distribution of Y.
However, if Y is also a standard normal random variable, then the sum X + Y will follow a normal distribution with mean 0 (mean of X) + 0 (mean of Y) = 0, and standard deviation √(1^2 + 1^2) = √2.
In this case, to find the probability that X + Y ≤ 0, we can use the standard normal distribution table or a statistical software to calculate the cumulative probability up to zero:
P(X + Y ≤ 0) = P(Z ≤ 0), where Z is a standard normal random variable with mean 0 and standard deviation √2.
Looking up the value in the standard normal distribution table or using a statistical software, we find that P(Z ≤ 0) is approximately 0.5.
Therefore, if Y is also a standard normal random variable, the probability that X + Y is less than or equal to zero is approximately 0.5.