00:01
Hello students in the given problem we need to show that probability of determinant of x greater than lambda is less than expected value of determinant of x to the power p divided by lambda to the power.
00:14
Here we are going to use markov's inequality which states that for any non -negative random variable y and any constant a greater than 0 we have probability of y greater than or equal to the expected value of y divided by a provided that a is greater than 0.
00:46
Applying markov's inequality to the random variable y is equal to determinant of x to the power p and a constant a equal to lambda we have probability of determinant of x to the power p greater than lambda to the power p less than or equal to expected value of determinant of x to the power p divided by lambda to the power p.
01:13
Since determinant of x greater than lambda is equivalent to determinant of x to the power p greater than or equal to lambda power p.
01:22
We can rewrite the equation as probability of determinant of x greater than lambda is less than or equal to expected value of determinant of x to the power p divided by lambda power p.
01:37
This proves the first part.
01:39
For the second part, we need to consider the sequence sn of a real valued random variable such that expected value of determinant of sn to the power 3 is less than or equal to n.
01:55
Here we need to prove that 1 upon n sn converges to 0 almost surely as n approaches to infinity.
02:05
To prove this, we need to show that for any epsilon greater than 0, the probability of the event of omega determinant of 1 upon n sn omega minus 0 is greater than or equal to epsilon converges to 0 as n approaches to infinity.
02:26
So first we assume that let epsilon greater than 0 be given and now we want to show that probability of determinant of 1 upon n sn minus 0 is greater than or equal to epsilon approaches to 0 as n converges to infinity...