This is an R Markdown document. Markdown is a simple formatting syntax for authoring HTML, PDF, and MS Word documents. For more details on using R Markdown see http://rmarkdown.rstudio.com.

When you click the **Knit** button a document will be generated that includes both content as well as the output of any embedded R code chunks within the document. You can embed an R code chunk like this:

`summary(cars)`

```
## speed dist
## Min. : 4.0 Min. : 2.00
## 1st Qu.:12.0 1st Qu.: 26.00
## Median :15.0 Median : 36.00
## Mean :15.4 Mean : 42.98
## 3rd Qu.:19.0 3rd Qu.: 56.00
## Max. :25.0 Max. :120.00
```

You can also embed plots, for example:

Note that the `echo = FALSE`

parameter was added to the code chunk to prevent printing of the R code that generated the plot.

**Problem 1**:

- A discrete random variable \(X\) has a probability mass function of the form \(P(X=x) =\frac{k}{2^x}\) for \(x= 1,2,3\) and zero otherwise, find \(k\).

We know that the \(\sum_{x=1}^{3} \frac{k}{2^x} = 1\). Therefore, we have that:

\[ \begin{aligned} \sum_{x=1}^{3} \frac{k}{2^x} &= \frac{k}{2^1} + \frac{k}{2^2} + \frac{k}{2^3} \\ &= \frac{k}{2} + \frac{k}{4} + \frac{k}{8} \\ &= \frac{7k}{8}\\ \Rightarrow k = \frac{8}{7} \\ \square \end{aligned} \]

- Can a function of the form \(f(x)=c(2^{-x}-0.5)\) for \(x= 0,1,2\) and zero otherwise be a probability mass function of a random variable?

\(f(0)=c(2^{-0}-0.5)= .50c \\\) \(f(1)=c(2^{-1}-0.5)= 0 \\\) \(f(2)=c(2^{-2}-0.5)=-.25c \\\) \(f(0) + f(1) + f(2) = .25 \\\) \(\Rightarrow c = 4 \\\)

However, we know that \(f(2)=4(2^{-2}-0.5)=-1\) Which is a contradiction since for a pdf \(f(x) \geq 0\). Thus \(f(x)\) is not a pdf \(\square\)

**Problem 2**: A function of the form \(f(t)=ct^{-c-1}I\{t >1\}\) for \(t \in (-\infty,\infty)\).

- If \(f(t)\) is a probability density function, find the value of \(c\).

We use the definition of a pdf to show that:

\[ \begin{aligned} \int_{-\infty}^{\infty} f(t) dt &= 1 \\ \int_{-\infty}^{\infty} f(t) dt &= \int_{-\infty}^{\infty} ct^{-c-1}I\{t >1\} dt \\ &= \int_{1}^{\infty} ct^{-c-1} dt \\ &= -t^{-c} \Big|_1^\infty \\ &= \frac{1}{\infty^c} + 1^{-c} \\ \Rightarrow c > 0 \end{aligned} \] We know this since if \(c = 0\) then we would get that \(f(t)=0 \Rightarrow \int_{-\infty}^{\infty} f(t)tx = \int_{-\infty}^{\infty} 0dt=0\) which is a contradiction of the definition of a pdf.

Furthermore if we have \(c < 0\) then we would get that \(\frac{1}{\infty^c} + 1^{-c} = \infty\), which also contradicts the definition of a pdf and that \(\int_{-\infty}^{\infty} f(t)dt = 1 \square\).

- Find the corresponding cumulative distribution function of \(f(t)\) in (1).

\[ \begin{aligned} P(T \leq t) &= F_T(t) \\ &= \int_{-\infty}^{t} f_T(x) dx \\ &= \int_{-\infty}^{t} cx^{-c-1}I\{x >1\} dx \\ &= \int_{1}^{t} cx^{-c-1} dx \\ &= -x^{-c} \Big|_1^t \\ &= -t^{-c} + 1^{-c} \\ &= (1-\frac{1}{t^c})I\{t >1\} \end{aligned} \]

**Problem 3**: Suppose \(f(t)\) and \(g(t)\) for \(t \in (-\infty,\infty)\) are probability density functions. Let \(a \geq 0\) and \(b \geq 0\) are two fixed constants satisfying \(a + b = 1\). Prove that \(af(t) + bg(t)\) is also a probability density function for \(t \in (-\infty,\infty)\).

To show that \(af(t) + bg(t)\) is a pdf we prove that \(af(t) + bg(t) \geq 0, \forall x\) and \(\int_{-\infty}^{\infty} af(t) + bg(t) dt = 1\).

- We first show that \(af(t) + bg(t) \geq 0\):

Assume that there exist a t such that \(af(t) + bg(t) < 0\). This implies that at least one of the two terms is the function are negative. From this we know that if \(af(t) < 0\) either \(a < 0\) or \(f(t) < 0\) which is a contradiction since we know that \(f(t) \geq 0\) (by the definition of a pdf) and \(a>0\) by the statement of the problem. If \(bg(t) < 0\) then either \(b < 0\) or \(g(t) < 0\) which is also a contradiction, since we know that \(g(t) \geq 0\) (by the definition of a pdf) and \(b>0\) by the statement of the problem. Therefore by contradiction, we know that \(af(t) + bg(t) \geq 0\)

- We show that \(\int_{-\infty}^{\infty} (af(t) + bg(t)) dx = 1\)

\[ \begin{aligned} \int_{-\infty}^{\infty} (af(t) + bg(t)) dt &= \int_{-\infty}^{\infty} af(t)dt + \int_{-\infty}^{\infty}bg(t) dx \\ &= a\int_{-\infty}^{\infty} f(t)dt + b\int_{-\infty}^{\infty}g(t) dx \\ &= a(1) + b(1), \space \space \space \space \space \space \space \space \text{since the Defintion of pdf we know that} \int_{-\infty}^{\infty} f(t)dt = 1, \int_{-\infty}^{\infty} g(t)dt = 1\\ &= a + b \\ &= 1 \\ \end{aligned} \]

From this we know that the definition of a pdf holds for \(af(t)+bg(t)\). \(\square\)