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Suppose $Z\sim N(\mu,1)$ and $V$ is independent of $Z$ with distribution $\chi^2_m$. Then $T=\frac{Z}{(V/m)^{1/2}}$ is said to have a noncentral $t$ distribution with noncentrality $\mu$ and $m$ degrees of freedom. I want to show that $$P(T\leq t)=2m\int_0^\infty \Phi(tw-\mu)f_m(mw^2)wdw$$ where $fm(w)$ is the $\chi^2_m$ density, and $\Phi$ is the normal distribution function.

I figured that $P(T<t)=E[P(T<t|V)]=\int P(T<t|V=v)f_mdv$ with change of variable $v=mw^2$, which I think should be somehow helpful here but I am stuck on how to actually derive the CDF of $T$.

Any helps would be appreciated!

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  • $\begingroup$ My best attempt: $$\begin{align} P(T \leq t \mid V = v) &= P\left( \dfrac{Z}{\sqrt{V/m}} \leq t\mid V = v\right) \\ &= P\left(\dfrac{Z}{\sqrt{v/m}} \leq t\right) \\ &= P(Z \leq t\sqrt{v/m}) \\ &= P(Z - \mu \leq t\sqrt{v/m}-\mu)\\ &= \Phi(t\sqrt{v/m} - \mu)\text{.} \end{align}$$ Hence you have $$P(T \leq t) = \int_{\mathbb{R}}\Phi(t\sqrt{v/m} - \mu)f_m(v)\text{ d}v\text{.}$$ Set $v = mw^2$ so that $\text{d}v = 2mw\text{ d}w$ and $\sqrt{v/m} = |w|$ and thus $$P(T \leq t) = 2m\int_{\mathbb{R}}\Phi(t|w| - \mu)f_m(mw^2)w\text{ d}w\text{.}$$ $\endgroup$
    – Clarinetist
    May 13, 2021 at 16:42

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