# CDF of noncentral $t$ distribution

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 $$P(T with change of variable $$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!

• 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 $$P(T \leq t) = 2m\int_{\mathbb{R}}\Phi(t|w| - \mu)f_m(mw^2)w\text{ d}w\text{.}$$ Commented May 13, 2021 at 16:42