Let U, V, and W be independent standard normal random variables (that is, independent normal random variables, each with mean 0 and variance 1), and let X=3U+4V and Y=U+W. Give a numerical answer for each part below. You may want to refer to the standard normal table .
What is the probability that X≥8?
P(X≥8)=
- unanswered
E[XY]=
- unanswered
var(X+Y)
To find the probability that X ≥ 8, we need to use the standard normal table.
First, let's find the mean and variance of X:
Since U and V are both standard normal random variables with mean 0 and variance 1, we can use the properties of linearity to calculate the mean and variance of X.
E[X] = E[3U + 4V] = 3E[U] + 4E[V] = 3(0) + 4(0) = 0
Var(X) = Var(3U + 4V) = 3^2Var(U) + 4^2Var(V) = 9(1) + 16(1) = 25
So, X is a normal random variable with mean 0 and variance 25.
To find the probability that X ≥ 8, we convert it to a standard normal random variable by standardizing it:
Z = (X - mean(X)) / sqrt(var(X)) = (8 - 0) / sqrt(25) = 8 / 5
Now, we can look up the probability using the standard normal table. The table provides the cumulative probability up to a certain value of Z. Since we want the probability that X ≥ 8, we need to find 1 - the cumulative probability up to Z = 8/5.
Using the standard normal table, we can find that the cumulative probability up to Z = 8/5 is approximately 0.8944. Therefore, P(X ≥ 8) = 1 - 0.8944 = 0.1056.
So, the probability that X ≥ 8 is approximately 0.1056.
For E[XY]:
E[XY] = E[(3U + 4V)(U + W)]
Since U, V, and W are independent, we can expand and compute the expected value term by term:
E[XY] = 3E[U^2] + 3E[UV] + 4E[VU] + 4E[V^2] + 3E[UW] + 4E[VW]
Since U, V, and W are standard normal random variables with mean 0 and variance 1, we know that E[U^2] = Var(U) + (E[U])^2 = 1 + 0^2 = 1.
Similarly, E[V^2] = E[W^2] = 1.
Since U, V, and W are independent, E[UV] = E[U]E[V] = (0)(0) = 0.
Using the same reasoning, E[VU] = E[V]E[U] = (0)(0) = 0.
E[UW] = E[U]E[W] = (0)(0) = 0.
E[VW] = E[V]E[W] = (0)(0) = 0.
Plugging in these values, we get:
E[XY] = 3(1) + 3(0) + 4(0) + 4(1) + 3(0) + 4(0) = 3 + 4 = 7.
So, E[XY] = 7.
For var(X + Y):
Since X and Y are independent standard normal random variables, their sum X + Y is also a normal random variable with mean E[X + Y] = E[X] + E[Y] and variance var(X + Y) = var(X) + var(Y).
We already calculated E[X] as 0, so we just need to calculate E[Y].
E[Y] = E[U + W] = E[U] + E[W] = 0 + 0 = 0.
Now, plug in the values:
var(X + Y) = var(X) + var(Y) = 25 + 1 = 26.
So, var(X + Y) = 26.
To summarize:
- P(X ≥ 8) is approximately 0.1056.
- E[XY] = 7.
- var(X + Y) = 26.