= Paper 2
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperia_2_2022.pdf

= 1A
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= Solution
{parent=1A}

<Differentiation under the integral sign> first gives
$$
I'(x)=\int_0^\pi \cos\theta\,e^{x\cos\theta}\,d\theta.
$$
On the other hand,
$$
\frac d{d\theta}\left(\sin\theta\,e^{x\cos\theta}\right)
=\cos\theta\,e^{x\cos\theta}
-x\sin^2\theta\,e^{x\cos\theta}.
$$
Its integral is zero because $\sin\theta$ vanishes at both endpoints. Hence
$$
\boxed{I'(x)=\int_0^\pi x\sin^2\theta\,e^{x\cos\theta}\,d\theta}.
$$

Differentiating once more,
$$
I''(x)=\int_0^\pi\cos^2\theta\,e^{x\cos\theta}\,d\theta.
$$
For $x\ne0$, the preceding identity and $\sin^2\theta+\cos^2\theta=1$ give
$$
I''+\frac1xI'-I
=\int_0^\pi(\cos^2\theta+\sin^2\theta-1)e^{x\cos\theta}\,d\theta=0.
$$
The equation extends through $x=0$ in its regular limiting form.

Solved by gpt-5.6-sol high.

= 2B
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= Solution
{parent=2B}

The characteristic polynomial of the <linear recurrence relation> is
$$
r^3-6r^2+12r-8=(r-2)^3.
$$
The repeated-root rule therefore gives
$$
x_n=(A+Bn+Cn^2)2^n.
$$
The condition $x_0=0$ gives $A=0$. The other two conditions give
$$
2(B+C)=4,\qquad
4(2B+4C)=24,
$$
so $B+C=2$ and $B+2C=3$. Thus $B=C=1$, and
$$
\boxed{x_n=n(n+1)2^n}.
$$

Solved by gpt-5.6-sol high.

= 3F
{parent=Paper 2}
{scope}
{title2=Probability}

= Solution
{parent=3F}

A function $f$ on an interval is a <convex function> if
$$
f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)
\qquad(0\leq t\leq1).
$$
<Jensen inequality> states that for an integrable random variable $X$, when all terms are defined,
$$
f(\mathbb E X)\leq\mathbb E[f(X)].
$$

The function $\phi(x)=x\log x$ is convex on $(0,\infty)$ because
$$
\phi''(x)=\frac1x>0.
$$
Applying Jensen's inequality to the uniform distribution on the positive numbers $x_1,\ldots,x_n$ gives
$$
\frac1n\sum_{i=1}^nx_i\log x_i
\geq
\left(\frac1n\sum_{i=1}^nx_i\right)
\log\left(\frac1n\sum_{i=1}^nx_i\right).
$$
Writing $S=\sum_i x_i$ and multiplying by $n/S$ yields
$$
\boxed{
\frac{\sum_{i=1}^nx_i\log x_i}{\sum_{i=1}^nx_i}
\geq\log\left(\frac{\sum_{i=1}^nx_i}{n}\right)}.
$$

Solved by gpt-5.6-sol high.

= 4F
{parent=Paper 2}
{scope}
{title2=Probability}

= Solution
{parent=4F}

Expanding around the <expected value> $\mu$,
$$
G(a)=\mathbb E[(X-\mu+\mu-a)^2]
=\mathbb E[(X-\mu)^2]+(\mu-a)^2,
$$
because $\mathbb E[X-\mu]=0$. Therefore
$$
\boxed{G(a)=\sigma^2+(\mu-a)^2\geq\sigma^2},
$$
with equality exactly when $a=\mu$.

For the absolute loss,
$$
H(a)=\int_{-\infty}^a(a-x)f(x)\,dx
+\int_a^\infty(x-a)f(x)\,dx.
$$
Leibniz differentiation gives, with $F(a)=\int_{-\infty}^af(x)\,dx$,
$$
H'(a)=F(a)-(1-F(a))=2F(a)-1.
$$
Thus $H$ decreases while $F(a)<1/2$ and increases while $F(a)>1/2$. It is minimized at any <median>, characterized in the continuous case by
$$
\boxed{\int_{-\infty}^af(x)\,dx=\frac12}.
$$

Solved by gpt-5.6-sol high.

= 5C
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= a
{parent=5c}
{scope}

= Solution
{parent=a}

For a normalized equation
$$
y''+P(x)y'+Q(x)y=0,
$$
$x_0$ is an ordinary point when $P,Q$ are analytic there. It is a regular singular point when $(x-x_0)P(x)$ and $(x-x_0)^2Q(x)$ are analytic there. These are the <ordinary point criterion for a second-order equation> and <regular singular point criterion for a second-order equation>.

Here
$$
P(x)=\frac1x-1,\qquad Q(x)=\frac{\lambda}{x},
$$
so $x=0$ is regular singular. Seek a <power-series solution of a differential equation>
$$
y=\sum_{n=0}^{\infty}a_nx^n.
$$
Equating the coefficient of $x^n$ gives
$$
(n+1)^2a_{n+1}+(\lambda-n)a_n=0,
$$
and hence
$$
a_{n+1}=\frac{n-\lambda}{(n+1)^2}a_n.
$$
Therefore
$$
\boxed{
a_n=a_0\frac{\prod_{j=0}^{n-1}(j-\lambda)}{(n!)^2}},
\qquad
y=a_0\sum_{n=0}^{\infty}
\frac{\prod_{j=0}^{n-1}(j-\lambda)}{(n!)^2}x^n.
$$
The series terminates exactly when $\lambda=N$ is a nonnegative integer: the factor with $j=N$ then makes $a_{N+1}=0$. Thus the polynomial solutions occur for
$$
\boxed{\lambda=0,1,2,\ldots}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=5c}
{scope}

= Solution
{parent=b}

After division by $x$, the coefficient of $y'$ is $P(x)=1/x-1$. The <Abel identity> for the <Wronskian> gives
$$
W'=-P(x)W=\left(1-\frac1x\right)W,
$$
so
$$
\boxed{W(x)=C\frac{e^x}{x}}.
$$

For $\lambda=1$, direct substitution confirms that $y_1=1-x$. <Reduction of order> gives a second solution proportional to
$$
y_2=(1-x)\int\frac{e^x}{x(1-x)^2}\,dx.
$$
To extract the requested coefficients, put
$$
y_2=(1-x)\log x+b_1x+b_2x^2+\cdots.
$$
For the differential operator
$$
L[y]=xy''+(1-x)y'+y,
$$
one finds
$$
L[(1-x)\log x]=x-3.
$$
The analytic correction $h=\sum_{n\geq1}b_nx^n$ must therefore satisfy $L[h]=3-x$. Its constant and linear coefficients give
$$
b_1=3,\qquad 4b_2=-1.
$$
Hence
$$
\boxed{
y_2=(1-x)\log x+3x-\frac14x^2+\cdots}.
$$

Solved by gpt-5.6-sol high.

= 6A
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= a
{parent=6a}
{scope}

= Solution
{parent=a}

The multivariable <chain rule> gives, at $(x,y)=(a+t\cos\gamma,b+t\sin\gamma)$,
$$
g'(t)=f_x\cos\gamma+f_y\sin\gamma,
$$
$$
g''(t)=f_{xx}\cos^2\gamma
+2f_{xy}\sin\gamma\cos\gamma
+f_{yy}\sin^2\gamma.
$$
Sufficient conditions for a strict local minimum at $t=0$ are $g'(0)=0$ and $g''(0)>0$.

At a <stationary point>, $g'(0)=0$ in every direction. Its second derivative is the quadratic form of the <Hessian matrix>. If
$$
f_{yy}>0,\qquad f_{xx}f_{yy}-f_{xy}^2>0,
$$
then completing the square gives
$$
f_{xx}u^2+2f_{xy}uv+f_{yy}v^2
=f_{yy}\left(v+\frac{f_{xy}}{f_{yy}}u\right)^2
+\frac{f_{xx}f_{yy}-f_{xy}^2}{f_{yy}}u^2>0
$$
for every nonzero direction $(u,v)$. The Hessian is therefore <positive-definite matrix>[positive definite], and the stationary point is a strict local minimum.

Solved by gpt-5.6-sol high.

= b
{parent=6a}
{scope}

= Solution
{parent=b}

The stationary equations are
$$
4x^3-6x+2y=0,\qquad 2x+2y=0.
$$
Thus $y=-x$ and $4x(x^2-2)=0$, giving
$$
(0,0),\qquad(\sqrt2,-\sqrt2),\qquad(-\sqrt2,\sqrt2).
$$
The <Hessian matrix> is
$$
H=\begin{pmatrix}12x^2-6&2\\2&2\end{pmatrix}.
$$
At either nonzero stationary point,
$$
H=\begin{pmatrix}18&2\\2&2\end{pmatrix},
\qquad \det H=32>0,
$$
so both are strict local minima.

Along a line through the origin,
$$
g(t)=f(t\cos\gamma,t\sin\gamma)
=t^2q(\gamma)+t^4\cos^4\gamma,
$$
where
$$
q(\gamma)=-3\cos^2\gamma
+2\sin\gamma\cos\gamma+\sin^2\gamma.
$$
When $\cos\gamma\ne0$ and $r=\tan\gamma$,
$$
q(\gamma)=\cos^2\gamma(r-1)(r+3).
$$
Consequently the origin is a local minimum along the line when
$$
\boxed{\tan\gamma\leq-3\quad\text{or}\quad\tan\gamma\geq1},
$$
including the equality cases because the positive quartic term then leads. It is also a minimum on the vertical line. It is a local maximum along the line when
$$
\boxed{-3<\tan\gamma<1}.
$$
In the first case the graph is locally bowl-shaped. In the second it initially bends downward from the origin, but the positive quartic term eventually turns it upward, producing the usual double-well profile.

Solved by gpt-5.6-sol high.

= 7C
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= a
{parent=7c}
{scope}

= Solution
{parent=a}

The <characteristic polynomial> is
$$
\det(A-\lambda I)=\lambda(\lambda-1).
$$
For $\lambda_1=0$ and $\lambda_2=1$, corresponding <eigenvectors> are
$$
u_1=\begin{pmatrix}2\\1\end{pmatrix},
\qquad
u_2=\begin{pmatrix}3\\1\end{pmatrix}.
$$
They are linearly independent, so the general homogeneous solution is
$$
\boxed{
z(t)=\alpha\begin{pmatrix}2\\1\end{pmatrix}
+\beta e^t\begin{pmatrix}3\\1\end{pmatrix}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=7c}
{scope}

= Solution
{parent=b}

Resolve the forcing in the eigenbasis:
$$
\begin{pmatrix}1\\a\end{pmatrix}
=(3a-1)u_1+(1-2a)u_2.
$$
In the zero-eigenvalue direction, a constant forcing produces the particular integral $(3a-1)t\,u_1$. In the unit-eigenvalue direction, a constant particular integral is $-(1-2a)u_2$. Thus a particular integral depends on time exactly when
$$
\boxed{a\ne\frac13}.
$$
Adding the complementary solution from part (a), the general solution is
$$
\boxed{
z(t)=\alpha u_1+\beta e^tu_2
+(3a-1)t\,u_1-(1-2a)u_2}.
$$
This is the eigenvector form of the usual <variation of parameters> calculation.

Solved by gpt-5.6-sol high.

= c
{parent=7c}
{scope}

= Solution
{parent=c}

For either eigenpair above, $\lambda_i\in\{0,1\}$ and hence $\lambda_i^n=\lambda_i$ for every positive integer $n$. Therefore
$$
\frac{d^n}{dt^n}(u_i e^{\lambda_it})
=\lambda_i^n u_i e^{\lambda_it}
=A u_i e^{\lambda_it},
$$
so both terms in the solution from part (a), and every linear combination of them, solve the higher-order system.

A system of two scalar differential equations of order $n$ has a $2n$-dimensional solution space. The displayed family supplies two independent solutions, so there must be
$$
\boxed{2n-2}
$$
further linearly independent solutions.

Solved by gpt-5.6-sol high.

= 8B
{parent=Paper 2}
{scope}
{title2=Differential Equations}

= a
{parent=8b}
{scope}

= Solution
{parent=a}

Write the vector field as
$$
\dot x=2x(4-x-y^2),\qquad \dot y=y(x-1).
$$
The <equilibrium points of a dynamical system> in the closed first quadrant are
$$
\boxed{(0,0),\quad(4,0),\quad(1,\sqrt3)}.
$$
The <jacobian matrix> is
$$
J(x,y)=
\begin{pmatrix}
8-4x-2y^2&-4xy\\
y&x-1
\end{pmatrix}.
$$

At $(0,0)$ its eigenvalues are $8,-1$, so this is a <saddle equilibrium>. Its unstable direction is the positive $x$-axis and its stable direction is the positive $y$-axis; to first order, nearby trajectories satisfy $\dot x=8x$, $\dot y=-y$.

At $(4,0)$ the eigenvalues are $-8,3$, so it is also a saddle. The $x$-axis is the stable direction, while trajectories entering the quadrant in the $y$ direction move away.

At $(1,\sqrt3)$,
$$
J=\begin{pmatrix}-2&-4\sqrt3\\ \sqrt3&0\end{pmatrix},
$$
whose eigenvalues are
$$
-1\mathbin{\pm}i\sqrt{11}.
$$
It is therefore a <stable spiral>. A point immediately to its right moves upward, so nearby trajectories spiral counterclockwise into the equilibrium. These eigendirections and the inward spiral give the requested local phase-portrait sketches.

Solved by gpt-5.6-sol high.

= b
{parent=8b}
{scope}

= Solution
{parent=b}

The only positive equilibrium solves $1-y=0$ and $x-1=0$, hence is
$$
\boxed{(x,y)=(1,1)}.
$$
Away from a nullcline,
$$
\frac{dy}{dx}
=\frac{3y(x-1)}{x(1-y)}.
$$
This is separable:
$$
\left(\frac1y-1\right)dy
=3\left(1-\frac1x\right)dx.
$$
Integration gives the <first integral>
$$
\log y-y=3x-3\log x+C.
$$
Thus one convenient conserved quantity is
$$
\boxed{E(x,y)=3x-3\log x+y-\log y}.
$$

Its gradient is
$$
\nabla E=\left(3-\frac3x,\,1-\frac1y\right),
$$
so $(1,1)$ is its only stationary point in the positive quadrant. Its <Hessian matrix> is
$$
\nabla^2E=
\begin{pmatrix}3/x^2&0\\0&1/y^2\end{pmatrix},
$$
which is positive definite everywhere. Hence $(1,1)$ is a strict, indeed global, minimum.

The nearby level curves of $E$ are closed curves surrounding this minimum. Since $E$ is constant along every trajectory, a solution starting on a sufficiently small nearby level set cannot leave the region bounded by a slightly larger level set. This proves <Lyapunov stability> of the equilibrium: solutions initially close to $(1,1)$ remain close for all time.

Solved by gpt-5.6-sol high.

= 9F
{parent=Paper 2}
{scope}
{title2=Probability}

= a
{parent=9f}
{scope}

= Solution
{parent=a}

Because $U$ and $V$ are bounded, their exponential series may be integrated term by term for every real $t$. Thus their <moment-generating functions> satisfy
$$
M_U(t)=\mathbb E[e^{tU}]
=\sum_{j=0}^{\infty}\frac{t^j}{j!}\mathbb E[U^j],
$$
and similarly for $V$. Equality of every moment gives equality term by term, so
$$
\boxed{M_U(t)=M_V(t)\quad\text{for every }t\in\mathbb R}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=9f}
{scope}

= Solution
{parent=b}

Normalization of the <Gaussian distribution> gives
$$
A=\frac1{\sqrt{2\pi}}.
$$
Completing the square,
$$
tx-\frac{x^2}{2}
=-\frac{(x-t)^2}{2}+\frac{t^2}{2}.
$$
Therefore
$$
M_X(t)=\frac1{\sqrt{2\pi}}
\int_{-\infty}^{\infty}e^{tx-x^2/2}\,dx
=e^{t^2/2},
$$
and hence
$$
\boxed{M_X(t)=e^{t^2/2}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=9f}
{scope}

= Solution
{parent=c}

The normalizing constant satisfies
$$
B^{-1}=\sum_{n\in\mathbb Z}e^{-n^2/2}.
$$
For an integer $k$,
$$
\mathbb E[e^{kY}]
=B\sum_{n\in\mathbb Z}e^{kn-n^2/2}
=Be^{k^2/2}\sum_{n\in\mathbb Z}e^{-(n-k)^2/2}.
$$
Translation by the integer $k$ merely permutes the summation indices, so the final sum is $B^{-1}$. Consequently
$$
\boxed{\mathbb E[e^{kY}]=e^{k^2/2}
=\mathbb E[e^{kX}]}.
$$
The last equality uses the <moment-generating function> of the standard normal variable from part (b).

Solved by gpt-5.6-sol high.

= d
{parent=9f}
{scope}

= Solution
{parent=d}

No. A standard counterexample comes from the <log-normal distribution>. Let
$$
f(x)=\frac1{x\sqrt{2\pi}}
\exp\left(-\frac{(\log x)^2}{2}\right),
\qquad x>0,
$$
and, for a fixed $0<|\varepsilon|\leq1$, let
$$
f_\varepsilon(x)
=f(x)\left[1+\varepsilon\sin(2\pi\log x)\right].
$$
This is a nonnegative density distinct from $f$. For every nonnegative integer $n$, substituting $z=\log x$ makes the difference of the $n$th moments proportional to
$$
\mathbb E[e^{nZ}\sin(2\pi Z)],
\qquad Z\sim N(0,1).
$$
It is the imaginary part of
$$
\mathbb E[e^{(n+2\pi i)Z}]
=\exp\left(\frac{(n+2\pi i)^2}{2}\right),
$$
which vanishes because its phase is $2\pi n$. The case $n=0$ also proves that $f_\varepsilon$ is normalized. Thus the two distributions have every finite moment equal but are different. Unbounded random variables need not be determined by their moments.

Solved by gpt-5.6-sol high.

= 10F
{parent=Paper 2}
{scope}
{title2=Probability}

= a
{parent=10f}
{scope}

= Solution
{parent=a}

The <probability-generating function> is
$$
G_X(s)=\sum_{m=1}^{\infty}\mathbb P(X=m)s^m.
$$
Differentiating $n$ times,
$$
G_X^{(n)}(s)
=\sum_{m=n}^{\infty}
\frac{m!}{(m-n)!}\mathbb P(X=m)s^{m-n}.
$$
At $s=0$, only the term $m=n$ remains. Hence
$$
\boxed{\mathbb P(X=n)=\frac{G_X^{(n)}(0)}{n!}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=10f}
{scope}

= Solution
{parent=b}

For $0\leq s\leq1$, independence gives
$$
G_{X+Y}(s)
=\mathbb E[s^{X+Y}]
=\mathbb E[s^Xs^Y]
=\mathbb E[s^X]\,\mathbb E[s^Y].
$$
Therefore
$$
\boxed{G_{X+Y}(s)=G_X(s)G_Y(s)}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=10f}
{scope}

= Solution
{parent=c}

For the <geometric distribution> on $\{1,2,\ldots\}$,
$$
\mathbb P(X=m)=p(1-p)^{m-1}.
$$
Summing the <geometric series>,
$$
G_X(s)
=\sum_{m=1}^{\infty}p(1-p)^{m-1}s^m
=\boxed{\frac{ps}{1-(1-p)s}}.
$$

Solved by gpt-5.6-sol high.

= d
{parent=10f}
{scope}

= Solution
{parent=d}

When $r$ red marbles remain, the next draw reduces their number with probability $r/n$. The waiting time $W_r$ for that reduction has a <geometric distribution> on $\{1,2,\ldots\}$ and therefore
$$
G_{W_r}(s)
=\frac{(r/n)s}{1-(1-r/n)s}
=\frac{rs}{n-(n-r)s}.
$$
The successive waiting times are independent, and
$$
T=W_n+W_{n-1}+\cdots+W_1.
$$
Using the product rule for probability-generating functions,
$$
\boxed{
G_T(s)=\prod_{r=1}^{n}
\frac{rs}{n-(n-r)s}}.
$$

Solved by gpt-5.6-sol high.

= 11F
{parent=Paper 2}
{scope}
{title2=Probability}

= a
{parent=11f}
{scope}

= Solution
{parent=a}

The number of heads has the <binomial distribution>. Choosing the $k$ head positions gives
$$
\boxed{\mathbb P(\text{exactly }k\text{ heads})
=\binom nk p^k(1-p)^{n-k}}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=11f}
{scope}

= Solution
{parent=b}

Conditioned on exactly $k$ heads, every $k$-element set of head positions is equally likely. There are $\binom nk$ such sets. A run containing all $k$ heads can begin at any of the positions
$$
1,2,\ldots,n-k+1,
$$
and each beginning determines one admissible set. Thus, for $k\geq1$,
$$
\boxed{
\mathbb P(\text{the }k\text{ heads are consecutive}\mid
\text{exactly }k\text{ heads})
=\frac{n-k+1}{\binom nk}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=11f}
{scope}

= Solution
{parent=c}

For the process to stop on toss $n$, the last toss must be the $k$th head, while the first $n-1$ tosses must contain exactly $k-1$ heads. Hence the <negative binomial distribution> gives
$$
\boxed{
\mathbb P(T=n)
=\binom{n-1}{k-1}p^k(1-p)^{n-k}},
\qquad n\geq k.
$$

Solved by gpt-5.6-sol high.

= d
{parent=11f}
{scope}

= Solution
{parent=d}

Let $E_j$ be the expected additional number of tosses when the current terminal run contains $j$ consecutive heads. Then $E_k=0$, and for $0\leq j<k$,
$$
E_j=1+pE_{j+1}+(1-p)E_0.
$$
For $0<p<1$, iterating this recurrence from $j=k-1$ down to $0$ gives
$$
E_0=\bigl(1+(1-p)E_0\bigr)
\frac{1-p^k}{1-p}.
$$
Solving,
$$
\boxed{
E_0=\frac{1-p^k}{(1-p)p^k}}.
$$
When $p=1$, the waiting time is deterministically $k$, which is also the continuous limit of this formula as $p\to1$.

Solved by gpt-5.6-sol high.

= 12F
{parent=Paper 2}
{scope}
{title2=Probability}

= a
{parent=12f}
{scope}

= Solution
{parent=a}

By <Tonelli theorem> for the nonnegative indicators,
$$
\mathbb E[N]
=\mathbb E\left[\sum_{n\geq1}\mathbf1_{A_n}\right]
=\sum_{n\geq1}\mathbb P(A_n).
$$
If this is finite, then $N$ is finite almost surely: alternatively, <Markov inequality> gives
$$
\mathbb P(N\geq m)\leq\frac{\mathbb E[N]}m\longrightarrow0.
$$
Since $\{N=\infty\}\subseteq\{N\geq m\}$ for every $m$,
$$
\boxed{\mathbb P(N=\infty)=0}.
$$
This is the first <Borel-Cantelli lemmas>[Borel-Cantelli lemma].

Solved by gpt-5.6-sol high.

= b
{parent=12f}
{scope}

= Solution
{parent=b}

Write $p_n=\mathbb P(A_n)$. Independence gives, first for finite partial counts and then by monotone convergence,
$$
\mathbb E[2^{-N}]
=\prod_{n\geq1}
\mathbb E[2^{-\mathbf1_{A_n}}]
=\prod_{n\geq1}\left(1-\frac{p_n}{2}\right).
$$
Using $1-x\leq e^{-x}$ factor by factor,
$$
\boxed{
\mathbb E[2^{-N}]
\leq\exp\left(-\frac12\sum_{n\geq1}p_n\right)
=e^{-\mathbb E[N]/2}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=12f}
{scope}

= Solution
{parent=c}

If $\sum_n\mathbb P(A_n)=\infty$, part (b) gives
$$
0\leq\mathbb E[2^{-N}]\leq e^{-\infty}=0.
$$
Thus $2^{-N}=0$ almost surely. By the convention in the question, this occurs exactly when $N=\infty$. Therefore
$$
\boxed{\mathbb P(N=\infty)=1}.
$$
This is the second <Borel-Cantelli lemmas>[Borel-Cantelli lemma] for independent events.

Solved by gpt-5.6-sol high.

= d
{parent=12f}
{scope}

= Solution
{parent=d}

Divide the keystrokes into disjoint blocks of five, and let $A_n$ be the event that block $n$ is exactly HELLO. The events are <independent events>, and
$$
\mathbb P(A_n)=26^{-5}.
$$
Consequently
$$
\sum_{n=1}^{\infty}\mathbb P(A_n)=\infty.
$$
Part (c) shows that infinitely many of these block events occur almost surely. Each occurrence is an occurrence of HELLO in the full typed sequence, so
$$
\boxed{\mathbb P(\text{HELLO appears infinitely often})=1}.
$$

Solved by gpt-5.6-sol high.