Can you explain this to me using an example?? ty
1. Given P(X) and P(Y) and P(X|Y)
How to show that X and Y are or are not independent?
How to find the joint probability of X and Y?
How to show the probability of X given Y?
How to show the probability of X or Y?
Given P(X), P(Y) and P(X|Y).
If X and Y are independent, then .
Also, X and Y are independent if and only if
.
To find the joint probability of X and Y, that is to find
, we
have,
.
Probability of X given Y.
Probability of X or Y,
.
Can you explain this to me using an example?? ty 1. Given P(X) and P(Y) and...
2. Let the joint probability density function of (X, Y) be given by {ay otherwise. 1 and 0 < y < 2, f(z,y) (a) [6 pts] Determine if X and Y are independent. (b) [6 pts] Find P{X+Y <1) B( (c) [6 pts) Find
2. Let the joint probability density function of (X, Y) be given by {ay otherwise. 1 and 0
Table 1 Joint PMF of X and Y in Example 5.1 x=01 X=1 | 1 Fig. 1 shows PXY()PXY( JointPMF ? 2 Fig. 1. Joint PMF of X and Y (Example 5.1). a. b. c. d. Find P(X-0,Y<1). Find the marginal PMFs of X and Y. Find P(Y-1X-0). Are X and Y independent?
1. Suppose you have two random variables, X and Y with joint distribution given by the following tables So, for example, the probability that Y o,x - 0 is 4, and the probability that Y (a) Find the marginal distributions (pmfs) of X and Y, denoted f(x),J(Y). (b) Find the conditional distribution (pmf) of Y give X, denoted f(YX). (c) Find the expected values of X and Y, EX), E(Y). (d) Find the variances of X and Y, Var(X),Var(Y). (e)...
The joint probability density function for continuous random variables X and Y is given below. f (x) = x + y, 0 < x < 1, 0 < y < 1 if; 0, degilse. (a) Show that this is a joint density function. (b) Find the marginal density of X . (c) Find the marginal density of Y . (d) Given Y = y find the conditional density of X . (e) P ( 1/2 < X < 1|Y =...
The joint probability density function for continuous random variables X and Y is given below. f (x) = x + y, 0 < x < 1, 0 < y < 1 if; 0, degilse. (a) Show that this is a joint density function. (b) Find the marginal density of X . (c) Find the marginal density of Y . (d) Given Y = y find the conditional density of X . (e) P ( 1/2 < X < 1|Y =...
The joint probability density function for continuous random variables X and Y is given below. f (x) = x + y, 0 < x < 1, 0 < y < 1 if; 0, degilse. (a) Show that this is a joint density function. (b) Find the marginal density of X . (c) Find the marginal density of Y . (d) Given Y = y find the conditional density of X . (e) P ( 1/2 < X < 1|Y =...
Assume that the joint density function of X and Y is given by f (x, y) = 4,0 < x < 2,0 < y = 2 and f (x, y) = 0 elsewhere. (a) Find P (X < 1, Y > 1). (b) Find the joint cumulative distribution function F(x, y) of the two random variables. Include all the regions. (c) Find P (X<Y). (d) Explain how the value of P (1 < X < 2,1 < Y < 2)...
Suppose the joint probability distribution of two binary random variables X and Y are given as follows. X/Y 1 0 1 2 1 4 0 + 1 (a) Show the marginal distribution of X. [2pts] (b) Find entropy H(Y). [2pts] (e) Find conditional entropy H(XY). (3pts] (d) Find mutual information I(X;Y). [3pts] 2 (e) Find joint entropy H(X,Y). (3pts) Note: The following three proofs are not related to the example in parts (a - e). You need to prove each...
The joint probability density function (pdf) of (X,Y ) is given by f(X,Y )(x,y) = 12/ 7 x(x + y), for 0 ≤ y ≤ 1, 0 ≤ x ≤ 1, 0, elsewhere. (a) Find the cumulative distribution function of (X,Y ). Make sure you derive expressions for the cdf in the regions • x < 0 or y < 0; • 0 ≤ x ≤ 1, 0 ≤ y ≤ 1; • x > 1, 0 ≤ y ≤...
4. Let X and Y be independent exponential random variables with pa- rameter ? 1. Given that X and Y are independent, their joint pdf is given by the product of the individual pdfs of X and Y, that is, fxy(x,y) = fx(x)fy(y) The joint pdf is defined over the same set of r-values and y-values that the individual pdfs were defined for. Using this information, calculate P(X - Y < t) where you can assume t is a positive...