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If $X_1 \sim N(0,1)$ and $X_2 \sim N(X_1^2, 1)$ then does $(X_1, X_2)$ follow a bivariate normal distribution?

My thinking is that $X_1 ^2$ is $\chi^2_1 $ since it's the square of a $N(0,1)$ random variable. And then since $X_2$ has a $\chi^2_1$ as its mean, the joint can't be normal?

Thanks for any help!

Robert B
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1 Answers1

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If it was bivariate normal, then the joint distribution would be elliptical. It is in fact U-shaped: high positive values of $X_2$ are associated with large magnitude values of $X_1$ whether positive or negative. As expected there is an underlying parabola.

The distribution of $X_2$ is not normal: it is right-skewed, with for example $E[X_2]=1$ but the median appears to be less than $0.75$.

As an illustration, here is a sample of $1000$ values with the R code

set.seed(2015)
n <- 1000
X_1 <- rnorm(n, mean=0)
X_2 <- rnorm(n, mean=x^2)
plot(X_2 ~ X_1)

enter image description here

Henry
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  • Thanks for your help Henry. Can the fact that it's not normal be deduced using the chi-squared argument I suggested? The reason I'm wondering is that I was asked this in an exam a few days ago and it's been puzzling me. We obviously didn't have access to R in the exam so I'm wondering how it could have been quickly solved?

    I also see your point about $X_2$ not being normal - the mean can obviously only take positive values, but we were told in the exam that it was...

     – Robert B May 22 '15 at 15:55
  • $X_2$ is not normally distributed in terms of its marginal distribution. So $X_1,X_2$ cannot have a bivariate normal distribution But, given $X_1$, it is true that $X_2$ is conditionally normally distributed. I illustrated the chart using R, but you should have been able to sketch it yourself. – Henry May 22 '15 at 18:05