Thread: Random variables

1.
X is a random variable which can take the following values: (This is a table, c/2 means c over 2)
x.............0....1.....2...3....4
P(X=x)....c/2...c....2c..c...2/c

State the conditions for P(X = x) to be a probability distribution, then find the value of c for which this will be the case.

2.
The random variable X represents the number of visitors a hospital patient has in a visiting period. A maximum of four visitors is permitted. C is found to have probability distribution:
x...........0.......1......2......3.......4
P(X=x)..0.1....0.2....0.4....0.2....0.1

Show that E[X] = 2 and E [X²] = 5.2, then write down the probability distribution of the random variable Y = (X-2)² and hence, or otherwise, find E[(X-2)²].

Originally Posted by Kiwigirl

1.
X is a random variable which can take the following values: (This is a table, c/2 means c over 2)
x.............0....1.....2...3....4
P(X=x)....c/2...c....2c..c...2/c

State the conditions for P(X = x) to be a probability distribution, then find the value of c for which this will be the case.

For P(X=x) to be a probability distribution we need:



Which in this case is:



So , but this has no roots so no value of
will make a probability distribution.

RonL

Originally Posted by Kiwigirl

2.
The random variable X represents the number of visitors a hospital patient has in a visiting period. A maximum of four visitors is permitted. C is found to have probability distribution:
x...........0.......1......2......3.......4
P(X=x)..0.1....0.2....0.4....0.2....0.1

Show that E[X] = 2 and E [X²] = 5.2, then write down the probability distribution of the random variable Y = (X-2)² and hence, or otherwise, find E[(X-2)²].

By definition the expectation of a discrete RV is:



So:



and:



The probability distribution of is:



So:



RonL