Given that X and Y are independent and normally distributed as N(0,3) and N(0,5) respectively, what is the expected value of (XY)^2?
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1Are $X$ and $Y$ independent? – Henry Jan 24 '23 at 11:52
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1@Henry yes, they are – spicy_springroll Jan 24 '23 at 12:16
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The trick here is to rewrite the expected value in terms of several rules that you have learned in class. Maybe the following rules ring a bell:
- $E[XY] = E[X]\cdot E[Y] \quad \text{if $X$ and $Y$ are independent}$
- $E[X^2] = \text{Var}[X]+ E[X]^2$
Sextus Empiricus
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1One more rule is needed https://stats.stackexchange.com/questions/52646/variance-of-product-of-multiple-independent-random-variables – Jacques Wainer Jan 24 '23 at 14:15
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1@Jacques That extra rule is not needed and applying it only complicates the solution. Another one that is used implicitly is that functions of independent variables (separately) remain independent. For this specific question I would point out that $E[X^2]=\operatorname{Var}(X)$ because $E[X]=0.$ That eliminates all probability calculations and allows one to write down the answer immediately, depending on how "N(0,3)" is interpreted (is its variance $3$ or $9$?). (+1) – whuber Jan 24 '23 at 15:10
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2@JacquesWainer: The more useful rule is that $(XY)^2 = X^2Y^2$. – Ilmari Karonen Jan 24 '23 at 19:53