Question

Hey @Zarkon, sorry to bother you, have a quick question about yesterday's problem

Answer 1

How did you figure out that it was (x^2-x+42)/12?

Answer 2

I got the individual expectations and they worked correctly, but I got my answer in terms of a sum over x

Answer 3

whats the question pls repost

Answer 4

I got expected value:\[E(Y|X=x)=\frac{1}{6}(x^2+\sum_{i=x+1}^{6}{i})\]@Zarkon simplified that to \[\frac{x^2-x+42}{12}\]Just wondering how.

Answer 5

For x = 1, 2, 3, 4, 5, 6, I get the same answer from both, but I was just wondering how he came up with that better looking result

Answer 6

Teach us @Zarkon !

Answer 7

\[\sum_{i=a}^{n}{i}=\sum_{i=1}^{n}{i}-\sum_{i=1}^{a-1}{i}\]

\[\sum_{i=1}^{n}{i}=\frac{n(n+1)}{2}\]

Answer 8

You can shift the index to simplify it.

Let n = i - x, so i = x+n

Then plug i=x+1 and i=6 into n = i - x to get
\[\large\sum_{i=x+1}^{6}{i} = \sum_{n=1}^{6-x}{(n+x)} \]
Then you can use the summation formulas like the one zarkon gave. Remember that x is just a constant, so its sum is pretty straightforward.

Answer 9

But it's easier to shift the index if you just think of it as subtracting x from the top and bottom limits of the sum, and then adding x inside the summation
\[\large\sum_{i=x+1}^{6}{i} =\sum_{i=1}^{6-x}{(i+x)} =\]Then you can break up the sum like so\[\large \sum_{i=1}^{6-x}{(i+x)} =\sum_{i=1}^{6-x}{i}+\sum_{i=1}^{6-x}{x}\]
Then just use summation formulas and remember that x is a constant.

Answer 10

Just use number 1 and 2 from here http://wonderfulworldofcalulus.weebly.com/uploads/2/8/2/3/28239103/1398375903.jpg

Answer 11

\[\frac{1}{6}(x^2+\sum_{i=x+1}^{6}{i})\]

\[=\frac{1}{6}(x^2+\sum_{i=1}^{6}{i}-\sum_{i=1}^{x}{i})\]

\[=\frac{1}{6}(x^2+\frac{6\cdot7}{2}-\frac{x(x+1)}{2})\]

\[=\frac{1}{12}(2x^2+42-x(x+1))\]

\[=\frac{1}{12}(x^2-x+42)\]