Variance of square of random variable

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To find the variance of the sum of the squares of two fair dice, the formula used is VAR(A^2 + B^2) = E(A^4) - E(A^2)^2 + E(B^4) - E(B^2)^2. The expected values for each die, E(A) and E(A^2), are calculated as 3.5 and 91/6, respectively. Since the dice are independent, the variance can be simplified using the property var(x+y) = var(x) + var(y). It's important to justify assumptions about independence and identical expectation values for both dice. The discussion emphasizes the need for clear reasoning in the calculations.
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Homework Statement



Lets say I roll 2 fair dice and take the sum of the square of each dice. What formula will be the variance?

Homework Equations


var(x)=e(x^2)-e(x)^2

The Attempt at a Solution


For dice A;
E(A)=3.5
E(A^2)=91/6
^ same for dice B.

VAR(A^2+B^2)=E(A^4)-E(A^2)^2+E(B^4)-E(B^2)^2 ?
 
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how did you get to that? I'd start from
var(A^2+B^2) = E((A^2+B^2)^2) - E((A^2+B^2))^2

and simplify from there, though you'll probably end up at a similar place you need to justify it
 
well since i am throwing two different dice and squaring them, this means that they are independent, shouldn't I be able to use var(x+y)=var(x)+var(y)?

VAR(A^2+B^2)=E(A^4)-E(A^2)^2+E(B^4)-E(B^2)^2
 
yep, you just need to mention your assumptions, you can also say the expectation values will be the same for A & B
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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