$E \left( \Phi \left( X+ a \sqrt{\frac{V}{n}} \right) \right)=P(T<\frac{a}{\sqrt{1+\sigma^2}})$ where $X\sim N(0 , \sigma^2)$ , $V \sim \chi^2_{n}$

Question

Prove that
$$E\left( \Phi \left(X+ a \sqrt{\frac{V}{n}} \right) \right) =P(T<\frac{a}{\sqrt{1+\sigma^2}})$$ where $X\sim N(0 , \sigma^2)$ , $V \sim \chi^2_{n}$ is chi-square distribution with $n$ degree of freedom , $X$ and $V$ are independent , $T$ has Student's t-distribution with $n$ degrees of freedom and $\Phi$ is c.d.f of the standard normal distribution, that is $\Phi(y)=\int_{-\infty}^{y}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}} \, dt$.

Answer 1

$$E\left( \Phi \left(X+ a \sqrt{\frac{V}{n}} \right) \right) =\int_0^{+\infty} \int_{-\infty}^{+\infty} \Phi \left(x+ a \sqrt{\frac{v}{n}} \right) f_X(x) f_V(v) dx dv $$

$$=\int_0^{+\infty} \int_{-\infty}^{+\infty} \left(\int_{-\infty}^{ x+ a \sqrt{\frac{v}{n}}} f_Y(y) dy \right) f_X(x) f_V(v) dx dv $$

$$=\iiint_{\{Y<X +a \sqrt{\frac{V}{n}} \}} f_Y(y) f_X(x) f_V(v) dx dv dy $$

Where $Y\sim Normal(0,1)$

$$=P(Y<X +a \sqrt{\frac{V}{n}} )=P( \frac{Y-X}{\sqrt{\frac{V}{n}}} <a)$$

so $Y-X\sim N(0,1+\sigma^2)$ and $ \frac{\frac{Y-X}{\sqrt{1+\sigma^2}}}{\sqrt{\frac{V}{n}}}\sim t_n$(Student's t-distribution)

$$=P( \frac{\frac{Y-X}{\sqrt{1+\sigma^2}}}{\sqrt{\frac{V}{n}}} <\frac{a}{\sqrt{1+\sigma^2}})$$

$$=P(T<\frac{a}{\sqrt{1+\sigma^2}})$$ where $T$ has Student's t-distribution with $n$ degrees of freedom.