Problem source

The probability density function $\f(x)$ of the random variable $X$ is given by
$$\f(x) = k\left[{\phi}(x) + {\lambda}\g(x)\right]$$
where ${\phi}(x)$ is the probability density function  of a normal variate with mean 0 and variance 1, $\lambda $ is a positive constant, and $\g(x)$ is a probability density function defined by
\[ \g(x)= \begin{cases}
1/\lambda  & \mbox{for $0 \le x \le {\lambda}$}\,;\\
             0& \mbox{otherwise} .
\end{cases}
\]
Find $\mu$, the mean of $X$, in terms of $\lambda$, and prove that $\sigma$, the standard deviation of $X$, satisfies.
$$\sigma^2 = \frac{\lambda^4 +4{\lambda}^3+12{\lambda}+12} 
         {12(1 + \lambda )^2}\;.$$ 
In  the case $\lambda=2$:
\begin{questionparts}
\item draw a sketch of the curve $y=\f(x)$;
\item express the cumulative distribution function of $X$ in terms of $\Phi(x)$, the cumulative distribution function corresponding to $\phi(x)$;
\item evaluate $\P(0 < X < \mu+2\sigma)$, given that 
$\Phi (\frac 23 + \frac23 \surd7)=0.9921$. 
\end{questionparts}

Solution source

\begin{align*}
&& 1 &= \int_{-\infty}^{\infty} f(x) \d x \\
&&&= k[1 + \lambda] \\
\Rightarrow && k &= \frac{1}{1+\lambda} \\
\\
&& \mu &= \int_{-\infty}^\infty x f(x) \d x \\
&&&= k \int_{-\infty}^\infty x \phi(x) \d x + k \lambda \int_{-\infty}^{\infty} x g(x) \d x \\
&&&= k \cdot 0 + k \lambda \cdot \frac{\lambda}{2} \\
&&&= \frac{\lambda^2}{2(1+\lambda)} \\
\\
&& \E[X^2] &= \int_{-\infty}^\infty x^2 f(x) \d x  \\
&&&= k \int_{-\infty}^\infty x^2 \phi(x) \d x + k \lambda \int_{-\infty}^{\infty} x^2 g(x) \d x \\
&&&= k \cdot 1 + k \lambda \int_0^{\lambda} \frac{x^2}{\lambda} \d \lambda \\
&&&= k + \frac{k \lambda^3}{3} \\
&&&= \frac{3+\lambda^3}{3(1+\lambda)} \\
&& \var[X] &= \frac{3+\lambda^3}{3(1+\lambda)} - \frac{\lambda^4}{4(1+\lambda)^2} \\
&&& = \frac{(3+\lambda^3)4(1+\lambda) - 3\lambda^4}{12(1+\lambda)^2} \\
&&&= \frac{\lambda^4+4\lambda^3+12\lambda + 12}{12(1+\lambda)^2}
\end{align*}

\begin{questionparts}
\item $\,$ 
\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/sqrt(2*pi)*exp(-(#1)^2/2)};
    % \def\functiong(#1){sin(deg(10*#1))*exp(-#1)};
    \def\xl{-4}; 
    \def\xu{4};
    \def\yl{-0.25}; 
    \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},
        curveC/.style={very thick, color=green!70!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);
    
    % Set up axes
    \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:0, samples=450] 
            plot ({\x},{1/(1+2)*\functionf(\x)});
        \draw[curveA, domain=2:\xu, samples=450] 
            plot ({\x},{1/(1+2)*\functionf(\x)});
        \draw[curveA, domain=0:2, samples=450] 
            plot ({\x},{1/(1+2)*\functionf(\x) + 1/(1+2)});
        
        % \draw[curveB, domain=\xl:\xu, samples=450] 
            % plot ({\x},{\functiong(\x)});
    \end{scope}
    

    \end{tikzpicture}
\end{center}

\item $\,$ \begin{align*} && \mathbb{P}(X \leq x) &= \int_{-\infty}^x f(x) \d x \\
&&&= \begin{cases} 
\frac13 \Phi(x) & \text{if } x < 0 \\
\frac13\Phi(x) + \frac13x & \text{if } 0 \leq x \leq 2 \\
\frac13 \Phi(x) + \frac23 & \text{if } 2 < x
\end{cases}
\end{align*}

When $\lambda = 2$, $\mu = \frac{4}{6} = \frac23$, $\sigma^2 = \frac{16+32+24+12}{12 \cdot 9} = \frac{7}{9}$, so $\mu + 2 \sigma = \frac23 + \frac{2\sqrt7}{3}>2$.

Therefore \begin{align*}
&& \P(0 < X < \mu + 2\sigma) &= \frac13 \Phi\left (\frac{2+2\sqrt{7}}{3} \right) + \frac23 - \Phi(0) \\
&&&= \tfrac13 \cdot 0.9921 +\tfrac23 - \tfrac12 \\
&&&= 0.4974
\end{align*}
\end{questionparts}