Why Is the Expected Value of cos(t)sin(t) Zero?

AI Thread Summary
The expected value of the product cos(t)sin(t) is zero due to the symmetry of the sine and cosine functions over the interval [0, 2π). The integral of sin(t)cos(t) over this interval evaluates to zero, confirming that E[cos(t)sin(t)] = 0. This result holds under the normalized Lebesgue measure, which is commonly used in probability spaces. The discussion clarifies the mathematical reasoning behind this conclusion. Understanding this property is essential for grasping concepts in probability and trigonometric integrals.
pamparana
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Hello,

Just came across this that:

E[cos(t)sin(t)] = 0

the expected value of the product of cos(t)sin(t) is 0. However, I am unable to convince myself that is the case. Can anyone help me understand why this is so?

Many thanks,

Luc
 
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I suppose the probability space is [0,2 \pi) with normalized Lebesgue measure? In that case, what it means is
<br /> \frac{1}{2\pi}\int_0^{2\pi} \sin(t) \cos(t)\,dt = 0<br />
 
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