Mathematics: What best describes the slope of a vertical line?

What best describes the slope of a vertical line?

Ответ:

poppy1173

18.10.2020

Mathematics

a) The probability that X < 1 but Y > 1 is $\approx$ 0.45

b) The probability that X > Y is $\approx$ 0.24

Step-by-step explanation:

Given a set of random variables X, Y, ... , the joint probability distribution for of them is a probability distribution that gives the probability that each of X,Y, ... variables are in any given interval or discrete set of values specified for that variable. For the case of two random variables X and Y and the given joint probability density function f(x,y) this is:

$P(x\in [x_0, x_1], y\in [y_0, y_1])=\int_{x_0}^{x_1}\int_{y_0}^{y_1}f(x,y)dydx$ (1)

In the exercise the joint probability function is given and the interval of the variables goes from $x, y \in (0, \infty)$ .

a) Applying the definition given in (1):

See the first attached figure for a visualization of the area we need to integrated. It is depicted with a grey rectangle, while the colorful area is a density plot of the joint probability density function.

$P(x1) = \int_{0}^{1}\int_{1}^{\infty} e^{-2x-y/2}\cdot \left(\frac{3}{4}e^{-x}+e^{-y/2} \right)dydx\\\boxed{P(x1) = 0.447\approx0.45}$

b) In this, case the integration limits for the random variable are given by the function y=x and notice (second figure attached) that the region where is below the function. Therefore $x\in[y, \infty)$ :

$P(xy) = \int_{0}^{\infty}\int_{y}^{\infty} e^{-2x-y/2}\cdot \left(\frac{3}{4}e^{-x}+e^{-y/2} \right)dydx\\P(xy) = \int_{0}^{\infty} \frac{1}{4} e^{-7 y/2} \cdot(1 + 2 e^{y/2})dy\\\boxed{P(xy) = 0.238 \approx 0.24}$

Suppose that we are waiting for two events a and b to occur. x = the time until event a occurs and y

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What best describes the slope of a vertical line?

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