Probability about standard normal random variables

We have $Z_1,Z_2,Z_3$ are independent standard normal random variables. Find

a) $\mathbb P(Z_1<Z_2+Z_3)$

b) $\mathrm{Var}(Z_1Z_2^2)$

c) $\mathbb P(Z_1/Z_2>1)$

d) $\mathbb P(Z_1^2>Z_2^2+Z_3^2)$

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Please read the FAQ on homework questions and revise your post accordingly. – cardinal Nov 12 '11 at 1:44

Re d), $(Z_2,Z_3)$ is a random point of the complex plane whose module and argument are independent, the argument being uniformly distributed and the module $R=\sqrt{Z_2^2+Z_3^2}$ having density $$u(r)=r\mathrm e^{-r^2/2}\cdot[r\gt0].$$ First application: For every $r\gt0$, $$\mathrm P(R^2\lt r^2)=\int\limits_0^ru(s)\mathrm ds=1-\mathrm e^{-r^2/2},$$ hence one is looking for $$A=\mathrm P(R^2\lt Z_1^2)=\mathrm E(1-\mathrm e^{-Z_1^2/2})=1-B\quad\mbox{with}\quad B=\mathrm E(\mathrm e^{-Z_1^2/2}).$$ Second application: $$C=\mathrm E(\mathrm e^{-R^2/2})=\int\limits_0^{+\infty}\mathrm e^{-r^2/2}\,u(r)\mathrm dr=\left.-(1/2)\mathrm e^{-r^2}\right|_0^{+\infty}=1/2.$$ On the other hand, $Z_2$ and $Z_3$ are i.i.d. and distributed like $Z_1$ hence $$C=\mathrm E(\mathrm e^{-Z_2^2/2}\mathrm e^{-Z_3^2/2})=\mathrm E(\mathrm e^{-Z_2^2/2})\cdot\mathrm E(\mathrm e^{-Z_2^2/2})=B^2.$$ Conclusion: $A=1-1/\sqrt2$.

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Many questions, from easy to not so easy! Almost all the questions deal with basic items that must be covered by whatever book you are using. Summaries, and further links, are, as usual, available at the suitable Wikipedia entry.

a) The answer should be obvious by symmetry. But for more complicated problems of this general type, with means not necessarily $0$, and not necessarily equal, note that a linear combination of independent normals has normal distribution, with means and variances calculated by using the linearity of each. In our case, let $U=Z_1-Z_2-Z_3$. This is normal mean $0$ variance $3$. Of course we don't need to know the variance to see that $P(U<0)=1/2$.

b) Recall that $\text{Var}(W)=E(W)-(E(W))^2$. So we need to calculate $E(Z_1^2Z_2^4)$ and $E(Z_1Z_2^2)$. Use the fact that if $U$ and $V$ are independent, $E(UV)=E(U)E(V)$. Since $E(Z_1)=0$, we can see that $E(Z_1Z_2^2)=0$. Calculating $E(Z_1^2Z_2^4)$ is more troublesome. Since the standard normal has mean $0$ and variance $1$, $E(Z_1^2)=1$. For $E(Z_2^4)$, you could set up the appropriate integral $\int_{-\infty}^\infty z^4 f_Z(x)\,dx$ where $f_Z(z)$ is the density function of the standard normal. Integrate by parts, using $u=z^3$ and $dv=zf_Z(z)\,dz$. This reduces the problem, and quickly you will find that the expectation is $3$.

Or else if you know about these things, you can pick off the answer from the moment generating function of the standard normal. Most easily, $E(Z^{2n})$ can be looked up from many sources.

c) Added Please see the comment by @Henry below, in which he shows that by symmetry the answer is clearly $1/4$.

To solve similar problems with possibly different numbers, one may need to do two-variable integration. Or else you can look up the distribution of the random variable $Z_1/Z_2$. It is the Cauchy distribution, density function $\frac{1}{\pi}\cdot \frac{1}{1+x^2}$. If you are allowed to use that, the rest is first year calculus, the answer is $\frac{1}{4}$.

d) Again, this is work if you need to do it from the definition. It is easiest to look up the distribution of $Z_1^2$ and $Z_2^2+Z_3^2$ (they are both of the $\chi^2$ type). Then you calculate a two-dimensional integral. Or else you can look up the distribution of $\sqrt{Z_2^2+Z_3^2}$ (Rayleigh?). This still leaves a two-dimensional integral. Or else you can look up the distribution of the ratio $\frac{Z_1^2}{Z_2^2+Z_3^2}$. This is one of the $F$-distribution family. There may be a way to use symmetry to bypass this type of calculation, but I have not thought of one.

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I am not sure (c) is so difficult for a continuous distribution symmetric about 0. For $Z_1/Z_2>1$ you need the signs to be the same and $|Z_1|> |Z_2|$, so it is obviously $1/4$. – Henry Nov 12 '11 at 3:06
@Henry: True, the answer is obvious by symmetry. I missed it, though I felt a bit uneasy when writing down $1/4$, since it seemed to be giving me a message. But I didn't hear it. – André Nicolas Nov 12 '11 at 3:35