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Homework Problem:

It is known that a for a standard normal random variable $X$, we have ${\bf E}[X^3]=0$, ${\bf E}[X^4]=3$, ${\bf E}[X^5]=0$, ${\bf E}[X^6]=15$. Find the correlation coefficient between $X$ and $X^3$.

Answer:

Since ${\bf E}[X]={\bf E}[X^3]=0$, we have ${\rm cov}(X,X^3)={\bf E}[X\cdot X^3]={\bf E}[X^4]=3$. Furthermore, since ${\rm var}(X)=1$ and ${\rm var}(X^3)={\bf E}[X^6]=15$, we obtain $\rho (X,X^3)=\frac{3}{\sqrt {1}\cdot \sqrt {15}} =\sqrt {3/5}.$

Interestingly, even though the random variables are strongly dependent (the value of one determines the value of the other), the value of the correlation coefficient is moderate.


My approach was to try to solve for ${\bf E}[X]$ by calculating the covariance between various powers of $X$, but I couldn't come up with a way to isolate either ${\bf E}[X]$ or ${\rm var}(X)$.

The answer doesn't explain this. How is it determined that ${\bf E}[X]={\bf E}[X^3]=0$ and that ${\rm var}(X)=1$?

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    $\begingroup$ A standard normal random variable has mean 0 and variance 1. $\endgroup$ Mar 27, 2014 at 20:20
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    $\begingroup$ The standard normal is symmetric about $0$, so if $E(X^k)$ exists, where $k$ is odd, we must have $E(X^k)=0$. $\endgroup$ Mar 27, 2014 at 20:21

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Yes, it being a standard normal implies that $EX=0$ and $VX=1$. But even without that knowledge you could have figured it out since

$$0=EX^3= E(X-\mu)^3 + 3 \mu E(X-\mu)^2 + 3 \mu^2 E(X-\mu) + \mu^3 = 3\mu\sigma^2 + \mu^3,$$

such that $\mu=0$ or $\mu^2=3\sigma^2$.

Further,

$$3=EX^4= E(X-\mu)^4 + 4 \mu E(X-\mu)^3 + 6 \mu^2 E(X-\mu)^2 + 4 \mu^3 E(X-\mu)+\mu^4 \\ = 3\sigma^4 + 6 \mu^2 \sigma^2 + \mu^4,$$

etcetera.

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