Ex.1

Show \(x^{2} + exp(x) + 2 x^{4} + 1\) is convex.

Solution:

\(f( \alpha x + \beta y) \leq \alpha f(x) + \beta f(y)\)

\((\alpha x + \beta y)^{2} + exp(\alpha x + \beta y) + 2 (\alpha x + \beta y)^{4} + 1 \leq \alpha (x^{2} + exp(x) + 2 x^{4} + 1) + \beta (y^{2} + exp(y) + 2 y^{4} + 1)\)

Using \(\alpha + \beta = 1\) we can simplify the inequality and rewrite it as

\(2 \alpha x^{4} + \alpha x^{2} + \alpha exp(x) + 2 \beta y^{4} + \beta y^{2} + \beta exp(y) + 1 - ((\alpha x + \beta y)^{2} + exp(\alpha x + \beta y) + 2 (\alpha x + \beta y)^{4} + 1) \geq 0\)

\(2 \alpha x^{4} + \alpha x^{2} + \alpha exp(x) + 2 \beta y^{4} + \beta y^{2} + \beta exp(y) - (\alpha x + \beta y)^{2} - exp(\alpha x - \beta y) - 2 (\alpha x + \beta y)^{4} \geq 0\)

The inequality is always true which shows that \(x^{2} + exp(x) + 2 x^{4} + 1\) is convex.

Ex.2

Show that the mean of the exponential distribution \(p(x) = \begin{cases} \lambda e^{- \lambda x} & x \geq 0 ( \lambda > 0)\\0 & x < 0\end{cases}\) is \(\mu = \frac{1}{ \lambda }\) and its variance is \(\sigma ^{2} = \frac{1}{ \lambda ^{2} }\)

Solution:

To find the mean of the exponential distribution solve \(\mu = E[X] = \int_0^ \infty x \lambda e^{- \lambda x}\)

Solve using integration by parts \(\int u v dx = u \int v dx - \int u' (\int v dx) dx\)

\([-x e^{- \lambda x}]_0^ \infty + \int_0^ \infty e^{- \lambda x} dx\) =

\([-x e^{- \lambda x}]_0^ \infty + [ -\frac{1}{ \lambda } e^{- \lambda x}]_0^ \infty\) =

\((0 - 0) + (0 + \frac{1}{ \lambda })\) = \(\frac{1}{ \lambda }\)

To find the variance of the exponential distribution solve \(\sigma ^{2} = Var[X] = E[ X^{2}] - E[X] ^{2}\)

\(E[ X^{2}] = \int_0^ \infty x^{2} \lambda e^{- \lambda x}\) which can be solved by again using integration by parts

\([-x^{2} e^{- \lambda x}]_0^ \infty + \int_0^ \infty 2xe^{- \lambda x} dx\) =

\([-x^{2} e^{- \lambda x}]_0^ \infty +[- \frac{2}{ \lambda } x e^{- \lambda x} dx]_0^ \infty + \frac{2}{ \lambda } \int_0^ \infty e^{- \lambda x} dx\) Again use integration by parts =

\([-x^{2} e^{- \lambda x}]_0^ \infty +[- \frac{2}{ \lambda } x e^{- \lambda x} dx]_0^ \infty + \frac{2}{ \lambda } [- \frac{1}{ \lambda } x e^{- \lambda x} dx]_0^ \infty\) =

\((\frac{2}{ \lambda })( \frac{1}{ \lambda })\) = \(\frac{2}{ \lambda ^{2}}\)

\(E[ X^{2}] - E[X] ^{2} = \frac{2}{\lambda ^{2}} - (\frac{1}{ \lambda }) ^{2} = \frac{2}{\lambda ^{2}} - \frac{1}{\lambda ^{2}} = \frac{1}{ \lambda ^{2}}\)

Ex.3

It is estimated that there is a typo in every 250 data entries in a database, assuming the number of typos can obey the Poisson distribution. For a given 1000 data entries, what is the probability of exactly 4 typos? What is the probability of no typo at all? Use R to draw 1000 samples with \(\lambda = 4\) and show their histogram.

Solution:

lambda <- 4
x1 <- 4
four_typos <- (lambda ^ x1 * exp(-lambda)) / factorial(x1)
four_typos
## [1] 0.1953668

The probability of exactly 4 typos is 19.54%.

lambda <- 4
x2 <- 0
zero_typos <- (lambda ^ x2 * exp(-lambda)) / factorial(x2)
zero_typos
## [1] 0.01831564

The probability of no typos at all is 1.83%.

# Built in r function to check answers
q1 <- dpois(4, lambda = 4)
q1
## [1] 0.1953668
q2 <- dpois(0, lambda = 4)
q2
## [1] 0.01831564
samples <- 1:1000
values <- rpois(1000, lambda = 4)
hist(values)