Expecting squares

Let the balls have numbers $a$ , $b$ , $c$ and $d$ . We know that $\frac{a}{4}+\frac{b}{4}+\frac{c}{4}+\frac{d}{4}=8$ . In other words, $a+b+c+d=32$ . We wish to maximize $\frac{a^2}{4}+\frac{b^2}{4}+\frac{c^2}{4}+\frac{d^2}{4}$ . Since $x^2$ increases faster at higher $x$ (i.e $x^2$ is concave up), we see that the operation that adds one to the largest number and subtracts one from the second largest increases the expected value. Therefore, the optimal case would have one of the numbers (say $a$ ) very large, and the others small. Thus we let $a=29$ , and the rest have value $1$ . This gives an expected value of $\frac{29^2+1^2+1^2+1^2}{4}=$ $\boxed{211}$

Let the integers written on the balls be $a,b,c,d$ . Now to convert it in simple terms, we have been given that $a+b+c+d=32$ and we want to maximize $a^2+b^2+c^2+d^2$ with $a,b,c,d$ all $>0$ Now we write $\small a=x+1 \\ b=y+1 \\ c=z+1 \\ d=t+1$ where $x,y,z,t$ are $\geq 0$ $$ Now we write $\small a^2+b^2+c^2+d^2 = (a+b+c+d)^2-2(ab+bc+ca+ad+bd+cd)$ So by substituting we now have $\small a^2+b^2+c^2+d^2 = x^2+y^2+z^2+t^2+2(x+y+z+t)+4$ We also know that $x+y+z+t=28$ . $$ So now the equation gets turned as follows: $\small a^2+b^2+c^2+d^2=(28)^2 - 2(xy+yz+xz+xt+yt+zt)+60$ Now all we want to do is to minimize $\small xy+yz+xz+xt+yt+zt$ . We can easily see that this gets minimized if any $3$ of $x,y,z,t$ are $0$ . So our final answer is $\small \max\{a^2+b^2+c^2+d^2\}=28^2+60 = 844$ The final answer thus is $\boxed{\frac{844}{4} = 211}$

Let the four numbers be a , b , c and d .The expectation of these four numbers is equal to their arithmetic mean ,i.e.

E[X] = \frac {Sum of all numbers}{Number of elements} .

\Rightarrow E[X] = \frac { a + b + c + d }{4} = 8

\Rightarrow a + b + c + d = 32

Now,E[X^2] = \frac {Sum of squares of each number}{Number of elements} For E[X^2] to be maximum , the sum of squares of the numbers should be maximum which is only possible if one of the numbers is as large as possible and the others are minimum . As the sum of the numbers is 32 , the largest possible value of one of the numbers is 29 while the value of the other three numbers is equal to 1 . Therefore , the maximum value of sum of squares of the numbers = 1^2 + 1^2 + 1^2 + {29}^2 = 844

Hence , the maximum value of E[X^2] = \frac {844}{4} = 211 .

Let Xi be the integer on the i th ball and pi be the probability the the random drawn ball is the i th ball then, E[X]=\displaystyle \sum {i=1}^4(Xi*pi)=8 we know that pi=1/4 so \displaystyle \sum {i=1}^4(Xi)=32 E[X^2]=\displaystyle \sum {i=1}^4(Xi^2*pi) =(\displaystyle \sum {i=1}^4(Xi^2))/4 since \displaystyle \sum {i=1}^4Xi=32 therefore maximum value of \displaystyle \sum {i=1}^4Xi^2=844 So (\displaystyle \sum_{i=1}^4*Xi^2)/4=211

we know $Var (X) = E(X^2)- [E(X)]^2$ mean = $(x_1+x_2+x_3+x_4)/4=8$ and variance is $[(x_1-8)^2+(x_2-8)^2+(x_3-8)^4+(x_4-8)^2]/4$ $E[x^2]=var(x)+[E(x)]^2$ $= [x_1^2+x_2^2+x_3^2+x_4^2+256-16x_1-16x_2-16x_3-16x_4]/4+64$ now var(x )of x is maximum if $x_1=x_2=x_3=1$ and $x_4=29$ $=[1+1+1+841+256-16-16-16-464]/4+64=147+64=211$

Let, x1,x2,x3,x4 be the four values What's given is: x1 + x2 + x3 + x4 = 32 --- eq(1)

To find the maximum value of (x1^2 + x2^2+x3^2+ x4^2) As all the values are integers, the maximum occurs when one of them is as large as possible and the other three are minimum i.e, x1 = x2 = x3 = 1 and x4 = 29 Maximum value of (x1^2 + x2^2+x3^2+ x4^2) = 3 * 1^2 + 29^2 = 844

Hence, E[X^2] = 844/4 = 211

Answer : 211

As we want to maximize $E[X^2]$ , We want one of the balls' values to be as high as posible. Thus, let's say three of the balls have 1s on them and the other one must have a 29, because: $E[X]=\frac {b_1+b_2+b_3+b_4}{4} \Rightarrow 8=\frac {1+1+1+b_4}{4} \Rightarrow b_4=29$ .

With these values, we get that: $E[X^2]= \frac {b^2_1+b^2_2+b^2_3+b^2_4}{4} = \frac {1^2+1^2+1^2+29^2}{4} = \frac {1+1+1+841}{4} = \frac {844}{4} = 221$

Let $w,x,y,z$ be the values of the balls. $E[X^2] = (w^2 + x^2 + y^2 + z^2)/4.$ If we have $x \geq y \geq 2$ then if the balls had values $w,x+1,y-1,z$ then the expected value is $w^2 + x^2 + y^2 + z^2 + 2 +2(x-y).$ This is greater than the expected value with $w,x,y,z.$ Thus, the maximum expected value will occur when three of $w,x,y,z$ are 1 and the other is $29.$ In this case, the expected value is $\frac{3 + (29)^2}{4} =211$

4 8=32 (1+1+1+29 29)/4=211