00:01
Hello students, today we will discuss about this question.
00:03
In this question we have given let x1, x2, x3, that will be i, i, d random variables with mean that is equal to 0 and finite variance that is sigma square.
00:21
Now let s n is equal to x1 plus up to plus xn and determine the limits below which precise the just justifications.
00:34
So here in the part a that is limit n tends to infinite probability of sn is greater than 0 .01 and part b that is limit n tends to infinite probability of sn is greater than or equals to 0 and in the part c limit n tends to infinite probability of s n is greater than or equals to minus 0 .01 n so here we need to to mine the limits.
01:03
So that is equal to all that equals to question mark.
01:07
So here first of all, we can say that the expectations and the variance of sn that will be calculated as s .e of s .n that is equals to e of x1 plus e of x2 plus e of xn, that is equal to we can write zero.
01:27
And here, variance of s n variance of s n that is equals to variance of x1 plus variance of x2 plus up to plus variance of xn that is equal to n sigma square now here for the part a our objective is to find the limit so therefore here we can write the limit n tends to infinite probability of s n is greater than or equals to 0 .0 1n that is equals to here we can write the limit n tends to infinite probability of s n minus e of s n divided by square root of variance of s n is greater than or equals to 0 .01 and e of s n divided by square root of variance of s n this will be suppose equation 1 now here we can say that the with the finite here we can write the limit n tends to infinite probability of a is less than or equals to s and minus nm u divided by sigma square root of n less than or equals to b that nearly equals to phi of b minus phi of a so now using this theorem in the equation one so that is equal to here from the equation one we can write limit n tends to infinite probability of s n is greater than or equals to 0 .01 n that is equals to we will get limit n tends to infinite probability of z is greater than or equals to 0 .01 square of n divided by sigma so therefore here we will get z that is the standard normal variant so that nearly equals to probability of z is greater than or equals to infinite so that is equals to 0.
03:25
So this will be the required answer for the part a.
03:28
Now for the part b here we can say that limit n tends to infinite probability of s n is greater than or equals to 0 that is equals to limit n tends to infinite probability of s n minus e of sn divided by square root of variance of s n is greater than or equals to zero minus e of sn divided by square root of variance of s .n.
03:56
That will be supposed equation 2...