Statistics and Fundamentals of Probabilistics Methods

By using Central Limit Theorem (CLT), calculate an approximation to P {ω ∈ Ω : 10 X ! Xi (ω) > 6} i=1 SOLUTION For mean value, we have: Z1 µ = E [Xi ] = 0  2 1 x 1 x dx = = 1−0 2 0 2 For variance we have:   2 σ = V ar(Xi ) = E Xi2 − (E [Xi ]) = 2 Z1 0  3 1 1 x 1 1 x2 dx − = − = 1−0 4 3 0 4 12 From CLT, we know that   P Zn = Xi − nµ √ σ n converges to N (0, 1). So let’s write the problem as function of Z10 (in a simplified way): ( P 10 X i=1 ) Xi > 6 ( 10 X ! 1 1 =P Xi − 10 · > 6 − 10 · 2 2 i=1 ( 10 ! ) X =P Xi − 5 > 1 ) i=1    10  P     Xi − 5  1  i=1 r =P >r  1 1     10 10  12 12 n √ o = P Z10 > 1. ⇒ P {ω ∈ Ω : 10 X i=1 1 ! Xi (ω) > 6} ≈ 13. If 10 fair dice are rolled, find by using CLT, the approximation probability that the sum obtained is between 30 and 40, inclusive. SOLUTION For mean value, we have: µ = E [Xi ] = 1+2+3+4+5+6 = 3. For given join distribution of the random vector (X, Y )  0.

find the distribution of X and Y. Are X and Y independent? SOLUTION To find the marginal distributions just calculate the sum over columns (for X) and over lines (for Y ): fX = (0. ⇒ fX = (0. fY = (0. SOLUTION For fX (x) we have to sum each column:  fX (x) =    1 1 1 1 1 1 1 1 1 + + a, + + b, + + c = a + ,b + ,c + 6 12 6 12 6 12 4 4 4 And for fY (y) we have to sum each row:  fY (y) =    1 1 1 1 1 1 1 1 + + , + + ,a + b + c = , ,a + b + c 6 6 6 12 12 12 2 4 We know that the sum of each distribution must be equal to one: X fX (x) = 1 ⇒ a + b + c + 3 1 =1⇒a+b+c= 4 4 For second distribution the result is the same. Now let’s multiply the probabilities density to get the matrix: 1 fX,Y (1, 1) = fX (1)fY (1) ⇒ = 6 fX,Y (2, 2) = fX (2)fY (2) ⇒   1 1 1 a+ · ⇒a= 4 2 12 1 = 12  b+ 1 4  · We known that a+b+c= 1 1 ⇒ a=b=c= 4 12 6 1 1 ⇒b= 4 12 7.

Suppose that X(Ω) = {1, 2}, Y (Ω) = {−1, 1} and random vector (X, Y ) has the join distribution as follows 1  6 1 6 1 3 1 3   Find the distribution of Z = X + Y. SOLUTION First of all, let’s find X and Y distributions:  fX (x) =  fY (y) = 1 1 1 1 + , + 6 3 6 3  1 1 1 1 + , + 6 6 3 3   1 1 , 2 2   1 2 , 3 3  = = To determine the Z = X + Y distribution, we need just to calculate the convolution of the previous distributions: fZ = fX ∗ fY or more intuitively: fZ (z < 0) = 0 fZ (z = 0) = fX (1)fY (−1) = 1 1 1 · = 2 3 6 fZ (z = 1) = fX (2)fY (−1) = 1 1 1 · = 2 3 6 fZ (z = 2) = fX (1)fY (1) = 1 2 1 · = 2 3 3 fZ (z = 3) = fX (2)fY (1) = 1 2 1 · = 2 3 3 fZ (z > 3) = 0 So:  fZ (z) = 7 1 1 1 1 , , , 6 6 3 3  Introduction to Mathematical Statistics Second List of Problems 1. Assume that for the characteristic X of some general population we have a simple sample (x1 , x2 ,. And now the empirical cummulative distribution function that need no parameters, just a point per point in crescent order, increasing 1/n = 0. per point.

To simplify the process, let’s sort the vector: X = (0. For a simple sample given in the task 1, compute: (a) the mean of simple sample x̄n = X̄(ω0 ) = 1 (X1 (ω0 ) + X2 (ω0 ) + · · · + Xn (ω0 )) n (b) the variance of simple sample n s2n = Sn2 (ω0 ) = 1X 2 (xi − x̄n ) n i=1 (c) the S–square pick value ŝ2n = Ŝn2 (ω0 ) = n S 2 (ω0 ) n−1 n SOLUTION In this problem we will apply the given definitions to calculate what is asked for the following vector: (x1 , x2 ,. xn ) = (1. Similarly, we find the column corresponding to 1 − α/2 in the table for lower critical values and reject the null hypothesis if the test statistic is less than the tabled value. So there we go. In table you can find at http://www3. med. unipmn. it/~magnani/pdf/t-Student. pdf, in the row corresponding to k = 15 you will find the value 2.

at critical values table in column α/2 = 0. As the distribution is symmetric, for lower side the critical value is −2.