 Home

Codex Wiki

List of universities

University of Oxford

Course of the University of Oxford

Undergraduate course of the University of Oxford

Physics course of the University of Oxford

Year 1 of the physics course of the University of Oxford

Mathematical methods II course of the University of Oxford

Past exam of the mathematical methods II course of the University of Oxford

2019 resit

Assume $\lambda>0$, as required for normalization of the <exponential distribution>. Its <cumulative distribution function> is the integral of its <probability density function> from $-\infty$ to $x$:
$$
\boxed{F_X(x)=\begin{cases}0,&x<0,\\1-e^{-\lambda x},&x\geq0.\end{cases}}
$$
For the <expected value>, <integration by parts> gives
$$
\begin{aligned}
\mathbb E[X]
&=\int_0^\infty x\lambda e^{-\lambda x}\,dx\\
&=\left[-xe^{-\lambda x}\right]_0^\infty+\int_0^\infty e^{-\lambda x}\,dx
=\frac1\lambda.
\end{aligned}
$$
Similarly,
$$
\mathbb E[X^2]=\left[-x^2e^{-\lambda x}\right]_0^\infty
+2\int_0^\infty xe^{-\lambda x}\,dx=\frac2{\lambda^2}.
$$
The endpoint terms vanish because exponential decay dominates each fixed power of $x$. Therefore
$$
\boxed{\mathbb E[X]=\frac1\lambda,\qquad
\operatorname{Var}(X)=\mathbb E[X^2]-\mathbb E[X]^2=\frac1{\lambda^2}.}
$$
The mean has the same units as $X$, while the <variance> has their square; $\lambda$ has reciprocal units.

Solution

Past exam of the mathematical methods II course of the University of Oxford / 2019 resit / 3 / Solution

Past exam of the mathematical methods II course of the University of Oxford / 2019 resit / 3 / Solution

 Ancestors (11)

 Discussion (0)

 Articles by others on the same topic (0)

Past exam of the mathematical methods II course of the University of Oxford / 2019 resit / 3 / Solution

 Ancestors (11)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)