Problem source

Let $X$ be a random variable with a  Laplace distribution, so that   its probability density function is given by
\[
\f(x) = \frac12 \e^{-\vert x \vert }\;, 
\text{  $-\infty < x < \infty $}. \tag{$*$}
\]
Sketch $\f(x)$. Show that its moment generating function 
${\rm M}_X(\theta)$  is given by ${\rm M}_X(\theta)= (1-\theta^2)^{-1}$ and hence find the variance of $X$.
A frog is jumping up and down, attempting to land on the same  spot each time. In fact, in each of $n$ successive jumps he always lands on a fixed straight line but when he lands from  the $i$th jump ($i=1\,,2\,,\ldots\,,n$) his displacement from the point from which he jumped is $X_i\,$cm, where $X_i$ has the distribution $(*)$. His displacement from his starting point after $n$ jumps is $Y\,$cm (so that $Y=\sum\limits_{i=1}^n X_i$).
Each jump is independent of the others. Obtain the moment generating function  for $Y/ \sqrt {2n}$  and, by considering its logarithm, show that this moment generating function tends to $\exp(\frac12\theta^2)$ as $n\to\infty$.
Given that $\exp(\frac12\theta^2)$ is the moment generating function of the standard Normal random variable, estimate the least number of jumps such that there is a $5\%$ chance that the frog lands 25 cm or more from his starting point.

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){0.5 * exp(-abs(#1))};
    \def\xl{-4}; 
    \def\xu{4};
    \def\yl{-0.2}; \def\yu{.75};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the reusable styles to keep code clean
    \tikzset{
        x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) grid (\xu,\yu);
    

    \draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
    % Define the bounding region with clip
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);

        \draw[curveA, domain=\xl:\xu, samples=151] plot(\x, {\functionf(\x)});
        \filldraw (0,0.5) circle (1pt) node[left] {$\frac12$};
    \end{scope}
        % Set up axes


    \end{tikzpicture}
\end{center}

\begin{align*}
&& M_X(\theta) &= \E \left [ e^{\theta X} \right] \\
&&&= \int_{-\infty}^{\infty} e^{\theta x} f(x) \d x \\
&&&= \int_{-\infty}^0 e^{\theta x}\frac12 e^{x} \d x+ \int_0^{\infty} e^{\theta x} \frac12 e^{-x} \d x \\
&&&= \frac12 \left [ \frac{1}{1+\theta}e^{(1+\theta)x} \right]_{-\infty}^0 +\frac12 \left [ \frac{1}{\theta-1}e^{(\theta-1)x} \right]_{0}^{\infty} \\
&&&= \frac12 \left (  \frac{1}{1+ \theta} + \frac{1}{1-\theta} \right) \\
&&&= \frac{1}{1-\theta^2} = (1-\theta^2)^{-1} \\
&&&= 1 + \theta^2 + \theta^4 + \cdots \\
\\
\Rightarrow && \E[X] &= 0 \\
&& \E[X^2] &= 2 \\
\Rightarrow && \var[X] &= 2
\end{align*}

\begin{align*}
&& M_{Y/\sqrt{2n}}(\theta) &= \E \left [ \exp \left ( \theta \frac{Y}{\sqrt{2n}} \right) \right] \\
&&&= \E \left [ \exp \left ( \frac{\theta }{\sqrt{2n}} \sum_{i=1}^n X_i  \right) \right] \\
&&&= \E \left [ \prod_{i=1}^n \exp \left ( \frac{\theta }{\sqrt{2n}} X_i  \right) \right] \\
&&&=  \prod_{i=1}^n \E \left [\exp \left ( \frac{\theta }{\sqrt{2n}} X_i  \right) \right] \tag{independence}\\
&&&=  \prod_{i=1}^n M_{X_i} \left ( \frac{\theta }{\sqrt{2n}} \right)\\
&&&=  \prod_{i=1}^n M_{X} \left ( \frac{\theta }{\sqrt{2n}} \right)\\
&&&=  M_{X} \left ( \frac{\theta }{\sqrt{2n}} \right)^n\\
&&&= \left (1 - \frac{\theta^2}{2n} \right)^{-n} \to \exp(\tfrac12 \theta^2)
\end{align*}

Given that $M_{Y/\sqrt{2n}} \to M_Z$ we assume that $Y/\sqrt{2n} \to Z$ or $Y/\sqrt{2n} \approx Z$.

\begin{align*}
&& 5\% &\approx \mathbb{P}(|Z| > 2) \\
&&&\approx \mathbb{P} \left (|Y| > 2\sqrt{2n} \right)
\end{align*}

So we wish to choose $n$ such that $2\sqrt{2n} = 25$ or $n = \frac{625}8 \approx 78$ so take $n = 79$