Suppose a line $T$ through $\mathbf{y}$ meets each $L_i$. Its intersection with $L_i$ is different from $\mathbf{y}$, so $T$ lies in ${\Pi }_i$. A nonzero direction vector $\mathbf{d}$ for $T$ must therefore satisfyA square matrix admitting a nonzero vector $\mathbf{d}$ with $M\mathbf{d}=0$ has zero determinant. Expanding givesThus the necessary condition isThe condition is not sufficient. Take $\mathbf{y}=(2,1,0)$, which is on none of the three original lines. It satisfies the condition, since $2+1-2=1$. The three plane equations becomeTheir common intersection is the single line $T=\{(s,1,0):s\in \mathbb{R}\}$. Any proposed line through $\mathbf{y}$ meeting all three original lines would have to be this $T$. But $T$ never meets $L_1=\{(s,0,1):s\in \mathbb{R}\}$: their second and third coordinates differ. The zero determinant guarantees a common direction for the three planes; it does not guarantee the required intersections. This is a parallel obstruction to a transversal of three skew lines.

The three lines have the convenient parametric equationsFor $L_1$, substitute $\mathbf{a}=\mathbf{i}$, $\mathbf{b}=\mathbf{j}$ in the plane through a point and a line given by a cross product formula. Its normal vector is $\mathbf{j}+\mathbf{i}\wedge \mathbf{y}=(0,1-y_3,y_2)$. Cyclically permuting the coordinates gives all three planes:Each normal vector is nonzero because $\mathbf{y}$ is not on the corresponding line. Thus these are genuine planes, and each is unique.

Apply the central limit theorem to the independent and identically distributed random variables $X_i^2$, whose mean and variance are $\theta$ and ${\theta }^2$. It gives the approximate standard normal distributionThe supplied standard normal quantile $c=1.96$ leaves probability $0.025$ in each tail. HenceSince $\theta >0$, this event is equivalent toFor $n>1.96^2$, equivalently integer $n\ge 4$, both denominators below are positive. Inverting the inequalities gives the approximate $95\%$ confidence intervalThe approximation is justified for large $n$; positivity at $n\ge 4$ by itself does not guarantee accurate coverage. This construction is a confidence interval from a relative normal pivot: the unknown scale remains inside the pivotal ratio until the inequalities are inverted. Its meaning is that the random interval covers the fixed statistical parameter $\theta$ with probability approximately $0.95$ under repeated sampling.

The recurrence relation at $k=3$ gives $I_5=2\theta I_3=2{\theta }^2$. ConsequentlyBy variance additivity for independent random variables, the variance of their sample mean is

Be careful about the index: the probability density function already contains a factor $x$, so $\mathbb{E}[X_i^r]=I_{r+1}$. The recurrence relation at $k=1$ yields $I_3=\theta I_1=\theta$. Therefore linearity of expectation givesThus $\hat{\theta }$ is an unbiased estimator.

For $I_1$, the change of variables $t=x^2/\theta$ givesFor $k\ge 1$, use $d(e^{-x^2/\theta })=-(2x/\theta )e^{-x^2/\theta }dx$ and integration by parts:The boundary term vanishes: $x^{k+1}\to 0$ at zero, and the exponential decay dominates the power at infinity. These estimates also establish convergence of the improper integrals used above.

For an observed random sample with all $x_i>0$, which happens almost surely, independence gives the likelihood functionIgnoring a term independent of $\theta$, the log-likelihood and its derivative areThis derivative is positive for $0<\theta <S/n$ and negative for $\theta >S/n$. Thus there is a unique global maximum, and the maximum-likelihood estimator is

Assuming $g$ is measurable, applying it separately to independent and identically distributed random variables preserves those properties. Thus $W_i=g(X_i)$ have common mean ${\mu }_g$ and variance ${\sigma }_g^2$. For ${\sigma }_g^2>0$, the central limit theorem givesHence, for large $n$, the approximate probability distribution isThe second argument is the variance. If ${\sigma }_g^2=0$, this sample mean equals ${\mu }_g$ almost surely, so the appropriate description is a point mass at ${\mu }_g$.

The central limit theorem states that if $W_1,W_2,\dots$ are independent and identically distributed random variables with finite mean $\mu$ and finite positive variance ${\sigma }^2$, thenHere the arrow means convergence in distribution: for every real $z$, the probability that the expression on the left is at most $z$ tends to $\Phi (z)$, the standard normal distribution function. If ${\sigma }^2=0$, the variables equal $\mu$ almost surely and their sample mean is exactly $\mu$.

The probability density function must integrate to one, soAlso, linearity of expectation and the mean and variance of the standard normal distribution giveUsing the probability density function to compute this expected value givesThese are the values $\Gamma (1/2)$ and $\Gamma (3/2)$ of the gamma function.

Differentiate the cumulative distribution function for $y>0$, using ${\Phi }^{\prime }(z)=(2\pi {)}^{-1/2}{e}^{-z^2/2}$ and $d\sqrt{2y}/dy=1/\sqrt{2y}$. The probability density function isIts value at zero can be assigned arbitrarily without changing any probability. Although this probability density function diverges as $y\downarrow 0$, $F_Y(0)=0$ and there is no point mass at zero. Equivalently, $Y$ has a gamma distribution with shape $1/2$ and rate one.

Since $Y=Z^2/2$ is nonnegative, its cumulative distribution function is zero for $y<0$. For $y\ge 0$,The last equality uses the symmetry of the standard normal distribution. Thus

The second moment isUsing the mean $\mathbb{E}[X]=1$, the variance is therefore

Normalization of the probability density function and the specified mean give two equations:Dividing the second by the first gives $2b/3=1$. ConsequentlyBoth are positive, as required for the stated probability density function.

Expand the square and use linearity of expectation, together with $\mathbb{E}[X]=\mu$ and $\mathbb{E}[1]=1$:This proof applies whenever the second moment is finite, and does not require the random variable to be continuous.

For a random variable with a continuous probability distribution and probability density function $f_X$, the mean and variance areA finite mean requires $\int |x|f_X(x)dx<\infty$. A finite variance additionally requires the second moment to be finite; without that condition the nonnegative integral for the variance may be infinite.

Computing the same expected value directly from the Poisson distribution instead givesFor each fixed positive integer $n$, the ratio test shows that this sum converges. This identity is Dobiński's formula. The case $n=0$ also gives $B(0)=1$ if $0^0$ is interpreted as one in this particular moment formula.

Substitute $X$ into the polynomial identity and apply linearity of expectation to its finite sum. The factorial moments of the Poisson distribution in part (a)(iv) are all one, givingThus its $n$th moment counts set partitions: it equals the $n$th Bell number.

Every assignment determines a unique set partition into its nonempty colour classes. If there are $k$ such blocks of a set partition, there are $A(n,k)$ choices for that set partition. For each fixed choice, assign distinct colours to the $k$ blocks of a set partition inways. Although the blocks of a set partition are unlabelled, they are distinguishable by their elements: ordering them temporarily, for instance by their least elements, counts each assignment exactly once. The falling factorial is zero when $k>x$, as it should be. Summing over the possible numbers of used colours provesHere $A(n,k)$ is the Stirling number of the second kind. Both sides are polynomials in $x$ and agree at infinitely many integers $x>2$, so the identity also holds for every real $x$, including $0,1,2$. This extension will justify substituting the random variable $X$ in part (iii).

Each of the $n$ distinct elements can receive any of the $x$ labelled colours. The multiplication principle therefore givescolour assignments; colours are allowed to remain unused.