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Given that $Z\sim N(0,1), Y \sim \chi^2_{v}$, and assuming that $Z, Y$ are independent, we define $W=\frac{Z}{\sqrt{Y}}$.

I aim to find $E(W)$ and $Var(W)$, with possible defining of $v$.

Finding $E(W)$ was a cinch; $E(W) = E(Z)E(\frac{1}{\sqrt{Y}}) = 0\cdot E(\frac{1}{\sqrt{Y}}) = 0.$

Finding the variance is a bit tricky, and I got up to this certain point:

$Var(W)=E(W^2)-[E(W)]^2=[Var(Z)+(E(Z))^2]E(\frac{1}{Y})=E(\frac{1}{Y}).$

From the obtained expected value above, how am I able to derive the variance?

kjetil b halvorsen
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raven
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    Does this answer your question? [How do we visualize the Decomposition of Variance formula $\text{Var}[y] = \text{Var}_x[\text{E}[y|x]] + \text{E}_x[\text{Var}[y|x]]$?](https://stats.stackexchange.com/questions/494421/how-do-we-visualize-the-decomposition-of-variance-formula-textvary-text) – Xi'an Nov 23 '20 at 06:18
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    Note that $Z$ is a $t$ random variable. – Xi'an Nov 23 '20 at 09:10
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    Wouldn't we divide $Y_1$ by its d.f.? Or do you refer to some scaled version of the t-distribution? – Christoph Hanck Nov 23 '20 at 09:53
  • Why mention $Y_2?$ If irrelevant, please remove it from the question. If relevant, please explain. – BruceET Nov 23 '20 at 20:13
  • Xi'an means W is a t random variable. – Michael R. Chernick Nov 28 '20 at 16:04

1 Answers1

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If $Z \sim \mathsf{Norm}(0,1)$ and, independently, $Y\sim \mathsf{Chisq}(5),$ then (by definition of the t distribution) $W^\prime = \frac{Z}{\sqrt{Y/5}} = \sqrt{5}\frac{Z}{\sqrt{Y}} = \sqrt{5}W \sim \mathsf{T}(5),$ which has $Var(W^\prime) = 5/(5-2) = 5/3.$

Thus $Var(W) = \frac{1}{5}\cdot\frac{5}{3} = \frac{1}{3}.$ Generalize.

Simulation in R: With a million iterations, one can expect two or three-place accuracy.

set.seed(2020)
m = 10^6;  df = 5
z = rnorm(m);  y = rchisq(m, df)
w = z/sqrt(y)
mean(w)
[1] 0.0003773159   # aprx E(W) = 0
var(w)
[1] 0.333333       # aprx Var(W) = 1/3
BruceET
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