Calculate expected value and variance of d, d = sqrt(x^2+y^2)

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SUMMARY

The discussion focuses on calculating the expected value (muD) and variance (sigD^2) of the stochastic variable D, defined as D = sqrt(X^2 + Y^2), where X and Y are normally distributed non-independent variables. The user successfully derives the variance using the relationship sigD^2 = muX^2 + sigX^2 + muY^2 + sigY^2 - muD^2. However, they struggle to compute the expected value muD, realizing that the assumption muD = sqrt(muX^2 + muY^2) is incorrect due to the non-independence of X and Y. The discussion highlights the application of Jensen's inequality to illustrate that E[sqrt(X^2 + Y^2)] is greater than sqrt(E[X^2] + E[Y^2]).

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tschoni
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I'm bad at stochastics so really glad for any help

Homework Statement



I have two normally distributed NON INDEPENDENT stochastic variables X~N(muX,sigX^2) and Y~N(muY,sigY^2)
A third variable D is defined as D = sqrt(X^2 + Y^2).
Since Y and X are stochastic D will also be stochastic.

Homework Equations



But how to calculate expected value muD and variance sigD^2 properly?

The Attempt at a Solution



Calculate the variance sigD^2
D = sqrt(X^2 + Y^2) (1)
D^2 = X^2 + Y^2 (2)
E[D^2] = E[X^2 + Y^2] = E[X^2] + E[Y^2] (3)
sigD^2 = E[(D - muD)^2] = E[D^2] - muD^2 (4)
Using (3) in (4)
sigD^2 = E[X^2] + E[Y^2] - muD^2 (5)
sigD^2 = E[X]^2 + Var[X] + E[Y]^2 + Var[Y] - muD^2 (5)
sigD^2 = muX^2 + sigX^2 + muY^2 +sigY^2 - muD^2 (6)
So far so good.
Calculate muD
At this point I thought I'm done.
But what about muD? I don't have it!
At first tempting to assume
muD = sqrt(muX^2+muY^2)
But I don't think it's true, since X and Y are not independent. And even if it is true, how to show it.
If I start out
muD = E[D] = E[sqrt(X^2 + Y^2)]
I don't manage to come to a solution.

Really appreciate any help
 
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I think that finding Ed (d = sqrt(X^2 + Y^2) ) is possible only numerically. Your hoped-for result Ed = sqrt(muX^2 + muY^2) is false: it is an example of the so-called "fallacy of averages". In (x,y) space the function f(x,y) = sqrt(x^2 + y^2) is *convex*, so E f(X,Y) >= f(EX,EY), and for a spread out distribution like yours (which straddles the origin) the inequality will be strict. Google "Jensens inequality" for more on this.

RGV
 
Last edited:
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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