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the ans to d 1st part..will it be 0??
since F(X) is integration of d values of X,the expectation of F(X) will be sum of expectation of the random variable till X..which is sum of zeros..
is it correct??


sir i got the answer 0 according to the following method...
var(x(N))
=var(x1/n+x2/n+......+xn/n)
=var(x1/n)+var(x2/n)+.......+var(xn/n)+2cov(x1/n,x2/n)+2cov(x2/n,x3/n)+....+2cov(xn/n,x1/n)
=n(sq)s/(sq)n+2np/(sq)n
=(sq)s/n+2p/n
lim as n tends to infinity var(x(n)) will therefore be 0.
sir please point out the mistake i'm making............
thank u

Administrator

Hi Sonal, mistake is quoted in the following step
=var(x1/n)+var(x2/n)+.......+var(xn/n)+"2cov(x1/n,x2/n)+2cov(x2/n,x3/n)+....+2cov(xn/n,x1/n)"
according to what you wrote (as quoted) there will be n covariance type terms but thats not the case and there are (sq(n)  n)/2 such terms.
For example when n = 4 then we have (sq(n)  n)/2 = 6 terms and not 4. Here is the elaboration:
=var(x1/n)+var(x2/n)+var(x3/n)+var(x4/n)+2cov(x1/n,x2/n)+2cov(x1/n,x3/n)+2cov(x1/n,x4/n)+2cov(x2/n,x3/n)+2cov(x2/n,x4/n)+2cov(x3/n,x4/n)

Administrator

Solutions are as follows:
Let X be a Normally distributed random variable with mean 0 and variance 1. Let F(.) be the cumulative distribution function of the variable X. Then the expectation of F(X) is
(c) 1/2
(Hint: F(X) has uniform distribution on (0, 1))
Consider any finite integer r ≥ 2. Then lim(x→0) f(r, x)/a(x) equals,
(where f(r, x) = log(e)(Σ(0, r) (x^k)); and a(x) = Σ(1, ∞) ((x^k)/k!))
(b) 1
Note: Σ(a, b) g(k) is summation of g(k) over values of k from a to b
x^k is x to the power k

Administrator

Next one:
Consider 5 boxes, each containing 6 balls labelled 1, 2, 3, 4, 5, 6. Suppose one ball is drawn from each of the boxes. Denote by b(i), the label of the ball drawn from the ith box, i = 1, 2, 3, 4, 5. Then the number of ways in which the balls can be chosen such that b(1) < b(2) < b(3) < b(4) < b(5) is
(a) 1,
(b) 2,
(c) 5,
(d) 6.


it should be option (b) 2


d no of ways are..
1 2 3 4 5
1 2 3 4 6
1 2 3 5 6
1 2 4 5 6
1 3 4 5 6
2 3 4 5 6
so d ans is 6..rite??

Administrator

That's right.
Consider 5 boxes, each containing 6 balls labelled 1, 2, 3, 4, 5, 6. Suppose one ball is drawn from each of the boxes. Denote by b(i), the label of the ball drawn from the ith box, i = 1, 2, 3, 4, 5. Then the number of ways in which the balls can be chosen such that b(1) < b(2) < b(3) < b(4) < b(5) is
(d) 6.


sir can u explain y F(X) has an uniform dist??

Administrator

First of all, F(X) take values in the interval (0, 1) because F(.) is a cumulative distribution function. So, for 0 < a < 1 we have:
P(F(X) < a) = P(X < Finv(a)) (Since F is strictly increasing)
= F(Finv(a)) = a
This implies that F(X) is distributed uniformly because Probability that F(X) take values less than a = a ∀ a ∈ (0, 1)

Administrator

The sum C(n+0, 0) + C(n+1, 1) +............C(n+m, m)equals
(a) C(n+m+1, n+m),
(b) (n+m+1)C(n+m, n+1),
(c) C(n+m+1, n),
(d) C(n+m+1, n+1).
Consider the following 2variable linear regression where the error e(i)’s are independently and identically distributed with mean 0 and variance 1; y(i) = a + b(x(i) − Mean(x)) + e(i), i = 1, 2, . . . , n. where Mean(x) = (x(1) + x(2) + ....x(n))/n
Let a^ and b^ be ordinary least squares estimates of a and b respectively.
Then the correlation coefficient between a^ and b^ is
(a) 1,
(b) 0,
(c) −1,
(d) 1/2.


the ans to d 2nd questn is
1


ans 1 is option (b) and 2nd answer is (c) 1


can u plz xplain d 1st ques??


actually m sorry, it shuld b option a


i try 2 give an explanation, expanding the binomial coefficients, we get Ln/Ln + L(n+1)/Ln + L(n+2)/Ln L2 ++ L(n+m)/Ln Lm ; whereLn means factorial n solving we get 1+ (n+1) + {(n+1)(n+2)/L2} ++[{(n+1)(n+2)(n+3)(n+m)}/ Lm] therefore we get option a as solution


@ priyanka
Hi.. i got till the second last part.. where you get the simplified form of the coefficients.. but how does that imply the ans as (a).. could you xplain..


The function f(x) = x(square_root(x) + square_root(x + 9)) is
(a) continuously differentiable at x = 0,
(b) continuous but not differentiable at x = 0,
(c) differentiable but the derivative is not continuous at x = 0,
(d) not differentiable at x = 0.
Note: square_root(g(x)) stands for square root of g(x)
Sir, cud u explain hw u got it as continuously differentiable at x = 0

Administrator

Find the derivative of the function. Its easy to see that it exists. Notice that derivative is also a function, check for its continuity.

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