The Ratio of Independent Standard Normals

Why a Ratio of Normal Variables Is Not Normal

Normal random variables are familiar, but taking their ratio can produce a distribution with very different behavior. In particular, even when both variables are centered at zero and have variance one, their ratio does not have a normal distribution. Its density is heavy-tailed: values far from zero remain appreciably more likely than a normal model would suggest. The key is to account for what happens when the denominator is close to zero.

Let \(X\) and \(Y\) be independent standard normal random variables, and define \(R=X/Y\) when \(Y\ne 0\). A standard normal variable has density \(\phi(x)=(1/\sqrt{2\pi})e^{-x^2/2}\). Since \(Y\) has a continuous density, \(P(Y=0)=0\); thus the ratio is defined with probability one. The derivation below finds the density of \(R\) directly from the joint density of \(X\) and \(Y\).

Definition: A standard Cauchy random variable has density \(f(r)=1/(\pi(1+r^2))\), for \(-\infty<r<\infty\). Its cumulative distribution function is \(F(r)=1/2+(1/\pi)\arctan(r)\).

Deriving the Ratio Density

Use \(R=X/Y\) as the variable of interest and retain \(V=Y\) as a second variable. Keeping \(V\) records the denominator, so the pair \((R,V)\) contains enough information to recover both original variables. The inverse transformation is \(X=RV\) and \(Y=V\). Its Jacobian determinant is \(V\), so the absolute Jacobian is \(|V|\).

The joint density of \(X\) and \(Y\) is the product of their standard normal densities, because they are independent. Substituting the inverse transformation and multiplying by the absolute Jacobian gives the joint density of \(R\) and \(V\). Integrating out \(V\) then gives the density of the ratio.

Specify the joint model.
Independence and standard normal marginals give \(f_{X,Y}(x,y)=(1/(2\pi))e^{-(x^2+y^2)/2}\) on the whole plane.

Transform and find the Jacobian.
Set \(R=X/Y\), \(V=Y\), so \(x=rv\), \(y=v\), and the absolute Jacobian is \(|v|\).

Integrate out the denominator.
Integrate the transformed joint density over all real \(v\) to obtain \(f_R(r)\).

f_{R,V}(r,v) =f_{X,Y}(rv,v)|v| =\frac{1}{2\pi}|v|\exp\left(-\frac{(1+r^2)v^2}{2}\right).

For fixed \(r\), write \(a=1+r^2\), which is strictly positive. The integral over \(v\) is even, so it is twice the integral over the positive half-line. With the substitution \(w=av^2/2\), the factor \(v\,dv\) becomes \(dw/a\). Therefore the integral is \(2/a\), as shown here:

\begin{aligned} f_R(r) &=\int_{-\infty}^{\infty} f_{R,V}(r,v)\,dv\\ &=\frac{1}{2\pi}\,2\int_0^\infty v e^{-av^2/2}\,dv\\ &=\frac{1}{\pi}\,\frac{1}{a}\int_0^\infty e^{-w}\,dw\\ &=\frac{1}{\pi(1+r^2)}. \end{aligned}

This is the standard Cauchy density. The transformation is valid away from \(v=0\); the excluded line \(Y=0\) has probability zero under the joint continuous density, so it does not affect the distribution. As a check, the density integrates to one because \(\int_{-\infty}^{\infty}(1+r^2)^{-1}\,dr=\pi\).

The calculation also explains why the denominator matters. For a fixed ratio \(r\), many pairs \((X,Y)=(rv,v)\) are possible. The transformed density accounts for all of them, with the factor \(|v|\) supplied by the change of variables. When \(|Y|\) is small, the ratio can have a large magnitude even if \(X\) is not unusually large; this contributes to the heavy tails.

Worked Example: Derive and Verify the Ratio Law

Let \(X\) and \(Y\) be independent standard normal variables, and let \(R=X/Y\). Find the density of \(R\), then verify it is a probability density. The method is a bivariate change of variables; the common mistake to avoid is forgetting the absolute value of the Jacobian.

Step 1: State the joint density. Independence gives \(f_{X,Y}(x,y)=(1/(2\pi))e^{-(x^2+y^2)/2}\), for all real \(x,y\).

Step 2: Transform. Take \(R=X/Y\) and \(V=Y\). Then \(x=rv\) and \(y=v\). The Jacobian determinant of \((x,y)\) with respect to \((r,v)\) is \(v\), so the density multiplier is \(|v|\), not \(v\).

Step 3: Find the marginal density. Substitution gives \(f_{R,V}(r,v)=(1/(2\pi))|v|e^{-(1+r^2)v^2/2}\). Integrating over \(v\) yields \(f_R(r)=1/(\pi(1+r^2))\), using \(\int_{-\infty}^{\infty}|v|e^{-av^2/2}\,dv=2/a\) for \(a>0\).

Step 4: Check total probability. The antiderivative of \(1/(1+r^2)\) is \(\arctan(r)\), so

\int_{-\infty}^{\infty} f_R(r)\,dr =\frac{1}{\pi}\left[\arctan(r)\right]_{-\infty}^{\infty} =\frac{1}{\pi}\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)=1.

The density is correctly normalized and is the standard Cauchy density. If the Jacobian were incorrectly taken as \(v\), integrating over negative and positive \(v\) would cancel contributions and could not produce a valid density. The absolute value is essential.

Using the Cumulative Distribution Function

Integrating the density from negative infinity to \(r\) gives the standard Cauchy cumulative distribution function. It is useful to retain the arctangent form when calculating interval probabilities or solving for a cutoff.

F_R(r)=P(R\le r) =\int_{-\infty}^{r}\frac{1}{\pi(1+t^2)}\,dt =\frac{1}{2}+\frac{1}{\pi}\arctan(r).

For any \(a<b\), subtract the endpoint cumulative probabilities: \(P(a<R<b)=F_R(b)-F_R(a)\). A single point has probability zero under this continuous density, so choices of strict or non-strict inequality at a finite endpoint do not change the answer. The distribution is symmetric about zero, which also gives \(F_R(-r)=1-F_R(r)\).

Worked Example: Find a Probability Between Two Ratios

For independent standard normal variables \(X\) and \(Y\), define \(R=X/Y\). Find \(P(1/\sqrt{3}<R<\sqrt{3})\). The method is to subtract cumulative probabilities, rather than treating the ratio as normal or using a normal table.

At the upper endpoint, \(\arctan(\sqrt{3})=\pi/3\). At the lower endpoint, \(\arctan(1/\sqrt{3})=\pi/6\). Apply the Cauchy cumulative distribution function:

\begin{aligned} P(1/\sqrt{3}<R<\sqrt{3}) &=F_R(\sqrt{3})-F_R(1/\sqrt{3})\\ &=\left(\frac12+\frac13\right)-\left(\frac12+\frac16\right)\\ &=\frac16. \end{aligned}

Thus the probability is \(1/6=0.1667\) to four decimal places. The endpoint terms \(1/2\) cancel. A frequent error is to use the standard normal cumulative distribution function because the numerator and denominator are normal; the distribution of their ratio is Cauchy, not normal.

Central Probability Cutoffs and Heavy Tails

The Cauchy cumulative distribution function can also be inverted. If \(q\) is a probability strictly between zero and one, the \(q\)-quantile \(r_q\) solves \(F_R(r_q)=q\). Rearranging gives \(r_q=\tan(\pi(q-1/2))\). For a symmetric central probability \(p\), the positive cutoff \(c\) satisfies \(P(-c<R<c)=p\), so \(c=\tan(\pi p/2)\).

P(-c<R<c)=\frac{2}{\pi}\arctan(c), \qquad c=\tan\left(\frac{\pi p}{2}\right) \quad\text{when }P(-c<R<c)=p.

Worked Example: Find the Central 80% Cutoff

Let \(R\) be the ratio of two independent standard normal variables. Find \(c>0\) such that \(P(-c<R<c)=0.80\). This is a central-quantile problem, so the probability is split equally between the two tails.

By symmetry, each tail has probability \((1-0.80)/2=0.10\). Thus \(F_R(c)=0.90\). Using the Cauchy cumulative distribution function,

\frac12+\frac{1}{\pi}\arctan(c)=0.90 \quad\Longrightarrow\quad \arctan(c)=0.40\pi=72^\circ.

Therefore \(c=\tan(72^\circ)\). Evaluating and rounding only at the final step gives \(c=3.0777\) to four decimal places. The interval is approximately \((-3.0777,3.0777)\), and substituting its endpoint back into the central-probability formula gives \(2(0.40\pi)/\pi=0.80\). The trap is to use \(0.80\) as the cumulative probability at \(c\); the upper endpoint has cumulative probability \(0.90\), because the remaining \(0.20\) is split between both tails.

Interpretation and Common Exam Traps

The Cauchy density has no finite mean and no finite variance. For example, the integral defining \(E[|R|]\) includes a tail proportional to \(\int_1^\infty r/(1+r^2)\,dr\), which diverges. Symmetry does not make the ordinary expectation zero: the positive and negative parts do not have finite expectations. Consequently, familiar summaries such as a mean and standard deviation are not appropriate for this distribution.

The result depends on the stated model: the variables are independent and each is standard normal. The proof uses their product joint density and the matching quadratic terms in the exponent. If the assumptions change, do not automatically apply the standard Cauchy density. In particular, dependence changes the joint density and therefore changes the transformation calculation. Also keep the ratio orientation consistent: this derivation uses \(X/Y\); for independent standard normals, swapping numerator and denominator gives the same distribution by symmetry of the joint model.

Wrong distribution: normal numerator and denominator do not imply a normal ratio. Use the standard Cauchy law under the stated independent standard normal model.
Wrong Jacobian: the factor is \(|v|\), not \(v\). A density transformation uses the absolute value of the determinant.
Wrong tail allocation: a central probability leaves equal probability in each tail. For central probability \(p\), the upper cutoff has cumulative probability \((1+p)/2\).
Wrong moment conclusion: symmetry alone does not establish a finite expectation. The Cauchy tails make the ordinary mean and variance undefined.

Key takeaway: If \(X\) and \(Y\) are independent standard normal variables, then \(X/Y\) has density \(1/(\pi(1+r^2))\). Its cumulative distribution function is \(1/2+\arctan(r)/\pi\), and its heavy tails mean it has neither a finite mean nor a finite variance.

Check Your Understanding

For each problem, let the numerator and denominator be independent standard normal variables, and use the ratio density derived above.

For \(R=X/Y\), calculate \(P(0<R<1)\) using the cumulative distribution function. (Answer: \(0.2500\))
Find \(P(-\sqrt{3}<R<\sqrt{3})\), showing how symmetry or the two endpoint cumulative probabilities determine the result. (Answer: \(2/3=0.6667\))
Calculate \(P(R>1/\sqrt{3})\). Express the arctangent contribution as a fraction of \(\pi\) before giving the decimal probability. (Answer: \(1/3=0.3333\))
Find \(c>0\) such that the central interval \((-c,c)\) contains probability \(0.50\). (Answer: \(c=1\))
Calculate the probability that the ratio lies outside \([-2,2]\), using symmetry and the Cauchy cumulative distribution function. (Answer: \(0.2952\))

What You'll Learn