I have two random variable $X=\mathcal N(\mu,\sigma^2)$ and $Y=\mathcal N(0,\sigma^2) $ independent to each other.

Now, $Z=X^2$ and $W=Y^2$, are chi-square random variable with first degree of freedom.

My aim is to determine $Prob(Z<W)$.

$f_X(x)=\frac {1} {\sqrt(2\pi\sigma^2)} exp^ {\frac {-(x-\mu)^2}{2\sigma^2} }$ and $f_Y(y)=\frac {1} {\sqrt(2\pi\sigma^2)} exp^{\frac {-y^2}{2\sigma^2}} $

For $W=Y^2$, pdf of chi-square distribution is $f_W(w)=\frac {1} {\sqrt(2\pi\sigma^2w)} exp^ \frac {-w} {2\sigma^2}$

However, for $Z=X^2$, I don't know how to find $f_Z(z)$. I know, we can use R.V. transformation $V=\frac {X-\mu} \sigma, $ but slightly confused.

Now, $prob(Z<W)=\int prob(Z<w)f_W(w)dw=\int F_Z(w)f_W(w)dw$. (Please verify this.)

Please help me in the derivation.

  • $\begingroup$ By symmetry the probability is $1/2$. $\endgroup$ – André Nicolas Jul 7 '15 at 18:33
  • $\begingroup$ You should revise your question because $Z$ is only chi-square distributed if $X$ is a standard normal random variable; here, $X$ has some general mean $\mu$, thus $X^2$ is not in general chi-square. $\endgroup$ – heropup Jul 7 '15 at 20:06
  • $\begingroup$ @d.k.o. Please reconsider your decision to delete your answer because it works even when $X$ and $Y$ are not normal random variables whereas my answer works only for normal random variables. You could edit your answer to emphasize this aspect.... $\endgroup$ – Dilip Sarwate Jul 8 '15 at 12:45


Show that $X+Y$ and $X-Y$ are independent $N(\mu,2\sigma^2)$ random variables. (This is easier to do than you might suspect). Then,

\begin{align} P\{X^2 < Y^2\} &= P\{X^2-Y^2 < 0\}\\ &= P\{(X+Y)(X-Y) < 0\}\\ &= P\{X+Y > 0, X-Y < 0\} + P\{X+Y < 0, X-Y > 0\}\\ &= P\{X+Y > 0\}P\{X-Y < 0\} + P\{X+Y < 0\}P\{X-Y > 0\} \end{align}

where the last step follows from the independence of $X+Y$ and $X-Y$. Note that all four probabilities on that last right side above can be expressed in terms of the standard cumulative probability distribution function $\Phi(\cdot)$ of the standard normal random variable.

  • $\begingroup$ Your solution is much simpler than I originally thought. Thanks!! $\endgroup$ – Kunal Jul 8 '15 at 6:39

After following Dilip Sarwate's solution


Assume $W=X+Y$ and $Z=X-Y$;

It can be shown that $W = \mathcal N(\mu, 2\sigma^2)$ and $Z = \mathcal N(\mu, 2\sigma^2)$

$P\{X+Y>0\}=P\{W>0\}=\int_{0}^{\infty} \frac{1}{\sqrt(2\pi*2\sigma^2)} exp^{\frac{-(w-\mu)^2}{2*2\sigma^2}} dw$

Now $m=\frac{w-u}{\sqrt2\sigma}$ $dm=\frac{dw}{\sqrt2 \sigma}$

$w \to 0 \ \ m \to- \frac{\mu}{\sqrt2 \sigma} $ and $w \to \infty \ \ m \to \infty $

Therefore, $P\{W>0\}=\int_{- \frac{\mu}{\sqrt2 \sigma}}^{\infty} \frac{1}{\sqrt(2\pi)} exp^{-\frac {m^2}{2}}dm$

$Q(z)=\int_z^\infty\frac{1}{\sqrt(2\pi)} exp^{-\frac {y^2}{2}}dy$

Therefore, $P\{X+Y>0\}=P\{W>0\}=Q(-\frac{\mu}{\sqrt2 \sigma})$

The same result applies to $P\{X-Y>0\}=P\{Z>0\}=Q(-\frac{\mu}{\sqrt2 \sigma})$

Similarly we can find

$P\{X+Y<0\}=P\{W<0\}=Q(\frac{\mu}{\sqrt2 \sigma})$ and $P\{X-Y<0\}=P\{Z<0\}=Q(\frac{\mu}{\sqrt2 \sigma})$

Therefore, $P\{X^2<Y^2\}=P\{X+Y>0\}P\{X-Y<0\}+P\{X+Y<0\}P\{X-Y>0\}$

$=Q(-\frac{\mu}{\sqrt2 \sigma})Q(\frac{\mu}{\sqrt2 \sigma})+Q(\frac{\mu}{\sqrt2 \sigma})Q(-\frac{\mu}{\sqrt2 \sigma})$

$=2*Q(-\frac{\mu}{\sqrt2 \sigma})Q(\frac{\mu}{\sqrt2 \sigma})$

But $Q(-z)=1-Q(z)$

Therefore, $P\{X^2<Y^2\}=2*{Q(\frac{\mu}{\sqrt2 \sigma})}\{1-Q(\frac{\mu}{\sqrt2 \sigma})\}$

Can anybody verify the proof?

  • $\begingroup$ Thanks for accepting my hints as the answer. Yes, what you have written above is correct, but once again, you are missing the forest while carefully cutting down multiple trees. Since $Z$ and $W$ are identically distributed, $P(W>0)$ and $P(Z<0)$ are complementary probabilities: their sum is $1$. Thus, $P(X^2<Y^2)$ is necessarily of the form $2p(1-p)$ and so we need find only one probability. It is a standard result that for $T \sim N(a,b^2)$, $$P(T\leq c) = \Phi\left(\frac{c-a}{b}\right) = Q\left(\frac{a-c}{b}\right)$$ but working out the details one more time by hand is OK too. $\endgroup$ – Dilip Sarwate Jan 23 '16 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.