Expressing the PDF as an integral

Question

If U and V are independent normal random variables, each having mean 0 and variance 1, and if $$W={U\over V}$$ Use a conditioning argument, or otherwise, so that for any $w ∈ \mathbb{R}$ express the quantity $F_W (w) = \mathbb{P} [W ≤ w]$ as an integral.

Would I be correct with the solution $F_W (w) = \mathbb{P} [W ≤ w] = \int_0^1{F_V (wv)dv}$?

Answer 1

$$F_W(w)=\mathbb P[W<w]=\iint_{u/v\le w}f_U(u)f_V(v)\,du\,dv$$ The inequality $u/v\le w$ is equivalent to $u\ge vw$ for $v<0$ and to $u\le vw$ for $v>0$. Split outer integral into two integrals: $$\tag{*}\label{*}F_W(w)=\int_{-\infty}^{0}\int^{\infty}_{vw}f_U(u)f_V(v)\,du\,dv+\int_{0}^{\infty}\int_{-\infty}^{vw}f_U(u)f_V(v)\, du \,dv$$ $$F_W(w)=\int_{-\infty}^{0}(1-F_U(vw))f_V(v)\,dv+\int_{0}^{\infty}F_U(vw)f_V(v)\, dv$$ We then can use the symmetry of standard normal distribution $F_U(x)=1-F_U(-x)$ and rewrite $$F_W(w)=\int_{-\infty}^{0}F_U(-vw)f_V(v)\,dv+\int_{0}^{\infty}F_U(vw)f_V(v)\, dv$$ Replacement $-v=\nu$ in the first integral leads to $$F_W(w)=\int^{\infty}_{0}F_U(\nu w)f_V(\nu)\,d\nu+\int_{0}^{\infty}F_U(vw)f_V(v)\, dv=2\int_{0}^{\infty}F_U(vw)f_V(v)\, dv$$

Let us also find pdf of $U/V$ in general case. Start with (\ref{*}) and change variables in both inner integrals: $u\mapsto t$ s.t. $u=tv$. Keep in mind that $v<0$ inside the first inner integral and $v>0$ in the second one. $$F_W(w)=\int_{-\infty}^{0}\int^{-\infty}_{w}f_U(tv)f_V(v)v\,dt\,dv+\int_{0}^{\infty}\int_{-\infty}^{w}f_U(tv)f_V(v)v\, dt \,dv=$$ $$\int_{-\infty}^{0}\int_{-\infty}^{w}f_U(tv)f_V(v)|v|\,dt\,dv+\int_{0}^{\infty}\int_{-\infty}^{w}f_U(tv)f_V(v)|v|\, dt \,dv = \int_{-\infty}^{\infty}\int_{-\infty}^{w}f_U(tv)f_V(v)|v|\, dt \,dv$$ Switch the order of integration by Tonelli theorem for non-negative functions and get $$F_W(w)= \int_{-\infty}^{w} \underbrace{\int_{-\infty}^{\infty}f_U(tv)f_V(v)|v|\, dv}_{f_{U/V}(t)} \,dt.$$

Answer 2

You would need the double integral: $$\iint_{u/v\le w}f_U(u)f_V(v)\text{d}(u,v)=\int_{-\infty}^{0}\int_{vw}^{\infty}f_U(u)f_V(v)\text{d}(u,v)+\int_{0}^{\infty}\int_{-\infty}^{vw}f_U(u)f_V(v)\text{d}(u,v)$$

Answer 3

By definition, and using the independence

$$F_{\frac UV}(w)=P\left(\frac UV <w\right)=\int_{-\infty}^{\infty}P\left(\frac Uv<w\ \mid\ V=v\right)f_V(v)dv=$$ $$=\int_{-\infty}^{\infty}P\left(\frac Uv<w\right)f_V(v)dv=$$ $$=\int_{-\infty}^{0}P\left(U>vw\right)f_V(v)dv+\int_{0}^{\infty}P\left(U<vw\right)f_V(v)dv=$$ $$=\int_{-\infty}^{0}(1-P\left(U\le vw\right))f_V(v)dv+\int_{0}^{\infty}P\left(U<vw\right)f_V(v)dv=$$ $$=\int_{-\infty}^0f_V(v)\ dv-\int_{-\infty}^{0}F_U(vw)f_V(v)dv+\int_{0}^{\infty}F_U(vw)f_V(v)dv=$$ $$=\int_{-\infty}^0f_V(v)\ dv-\int_{-\infty}^{0}\int_{-\infty}^{vw}f_U(x)\ dxf_V(v)dv+\int_{0}^{\infty}\int_{-\infty}^{vw}f_U(x)\ dxf_V(v)dv.$$

In order to get the pdf of the ratio we take the derivative of the expression above with respect to $w$.

$$f_{\frac UV}(w)=\frac d{dw}F_{\frac UV}(w)=$$ $$=\frac d{dw}\left[-\int_{-\infty}^{0}\int_{-\infty}^{vw}f_U(x)\ dxf_V(v)dv+\int_{0}^{\infty}\int_{-\infty}^{vw}f_U(x)\ dxf_V(v)dv\right]=$$

$$=\int_{-\infty}^{0}\frac d{dw}\left[-\int_{-\infty}^{vw}f_U(x)\ dx\right]f_V(v)dv+\int_{0}^{\infty}\frac d{dw}\left[\int_{-\infty}^{vw}f_U(x)\ dx\right]f_V(v)dv=$$ $$=\int_{-\infty}^{0}(-v)f_U(vw)f_V(v)dv+\int_{0}^{\infty}vf_U(vw)f_V(v)dv=$$ $$=\int_{-\infty}^{\infty}\mid v\mid f_U(vw)f_V(v)dv.$$

... and this is independent from the normality of the distributions involved. The only restriction is that the pdf of $V$ exist. ($P(V=0)=0$ is important.)

If we take two independent standard normal variables then the integral to be evaluated becomes

$$f_{\frac UV}(w)=\frac1{2\pi}\int_{-\infty}^{\infty}\mid v\mid e^{-\frac12v^2(1+w^2)}\ dv=\frac1{\pi}\frac1{1+w^2}.$$ As it is known: the ratio distribution of two independent standard normal random variables is the simple form of the Cauchy distribution gained above.