A network source sends a sequence of zeros and ones, $X_1, X_2, ...$ with $X_i$(iid) Bernoulli with $p = P(X_i = 1), 0 < p < 1$.
Due to disturbances the received sequence is $Y_1, Y_2, ...$ with $Y_i = X_i + U_i$. The $U_i$ are (iid) $N(0, \sigma^2)$ with $\sigma^2 = \frac{1}{3}$.
If $Y_i > \frac{1}{2}$ the received value is interpreted as $X_i = 1$ and $0$ otherwise. Ie. the $i$th signal sent if the received signal is greater than $\frac{1}{2}$, and $0$ otherwise.
I am trying to -
- Find $P(Y_i > \frac{1}{2}$ as a function of $p$.
- Estimate $p$ if $162$ of $424$ observations at the receiver are $1$s and the rest are $0$.
Attempt
1.
$P(Y > \frac{1}{2})$
$= 1 - P(Y \le \frac{1}{2})$
$= 1 - P(X_i + U_i \le \frac{1}{2})$
$\bigg[X_i + U_i \le \frac{1}{2}$ only when $X_i = 0$ and $U_i \le \frac{1}{2}\bigg]$
$= 1 - P(X_i = 0, U_i \le \frac{1}{2})$
$= 1 - [P(X_i = 0)P(U_i \le \frac{1}{2})]$ as $X_i, U_i$ independent.
$= 1 - [(1 - p)P(\frac{U_i - 0}{\frac{1}{\sqrt{3}}} \le \frac{\frac{1}{2}}{\frac{1}{\sqrt{3}}})]$
$= 1 - [(1 - p)P(Z \le \frac{\sqrt{3}}{2})]$
$= 1 - [(1 - p)(0.8078)]$
$= 1 - [0.8078 - 0.8078p]$
$0.1922 + 0.8078p$
Is that correct?
2.
I am really not sure about this part. We have $162$ success from $424$ observations. From part 1. we have that the probability of a success is $\theta = P(Y_i > \frac{1}{2}) = 0.1922 + 0.8078p$. So it seems to me that we are trying to find a maximum likelihood estimate for $p$? It seems we have binomial distribution with $n = 424, x = 162$ and $\theta = 0.1922 + 0.8078p$.
So $L(\theta|162) = {424 \choose 162}\theta^{162}(1-\theta)^{262}$
We can disregard the constant ${424 \choose 162}$ as it won't affect the final answer.
Then $ln [L(\theta|162)] = ln [\theta^{162}(1-\theta)^{262}]$
$= 162 ln(\theta) + 262 ln(1-\theta)$
Differentiating w.r.t. $\theta$ and setting the result = $0$ gives
$\frac{162}{\theta} - \frac{262}{\theta} = 0$
Obviously something has gone wrong here as this equation is invalid. Is my approach correct? What should I do next?