Looks like a badly encoded/decoded symbol. It's supposed to be a minus sign, so you're asked to find the expectation of 2<em>X </em>² - <em>Y</em>.
If you don't know how <em>X</em> or <em>Y</em> are distributed, but you know E[<em>X</em> ²] and E[<em>Y</em>], then it's as simple as distributing the expectation over the sum:
E[2<em>X </em>² - <em>Y</em>] = 2 E[<em>X </em>²] - E[<em>Y</em>]
Or, if you're given the expectation and variance of <em>X</em>, you have
Var[<em>X</em>] = E[<em>X</em> ²] - E[<em>X</em>]²
→ E[2<em>X </em>² - <em>Y</em>] = 2 (Var[<em>X</em>] + E[<em>X</em>]²) - E[<em>Y</em>]
Otherwise, you may be given the density function, or joint density, in which case you can determine the expectations by computing an integral or sum.
(-1,-1)
I recommend photomath, it helps with all equations!
Answer:
(48 m^12)/n^8
Step-by-step explanation:
Simplify the following:
3 ((2 m^3)/(n^2))^4
Multiply each exponent in (2 m^3)/(n^2) by 4:
3×2^4 m^(4×3) n^(-2×4)
4 (-2) = -8:
3×2^4 m^(4×3) n^(-8)
4×3 = 12:
(3×2^4 m^12)/(n^8)
2^4 = (2^2)^2:
(3 (2^2)^2 m^12)/(n^8)
2^2 = 4:
(3×4^2 m^12)/(n^8)
4^2 = 16:
(3×16 m^12)/(n^8)
3×16 = 48:
Answer: (48 m^12)/n^8
To find the sum, add. To find the difference, subtract. It's simple terminology in mathematics. Hope I helped!
