Apologies for the formatting. I didn't really know how to format this.
Let $_1,\ldots, _n$ be independent and identically distributed (i.i.d.) random variables with distribution function : $$P(Y_i \le y \mid \alpha,\beta) = \begin{cases}0 \quad \text{if } y < 0 \\ (y/\beta)^\alpha \quad \text {if } 0 \le y \le \beta \\ 1 \quad \text{ if } y > \beta\end{cases} $$
where the parameters $\alpha$ and $\beta$ are positive.
Find the Maximum Likelihood Estimators (MLEs) of $\alpha$ and $\beta$.
I integrated the original function. to get $y^{\alpha+1}/ [(\beta^\alpha)(\alpha+1)]$ I take the likelihood function and then take its logarithm to get
$-\alpha n\times \ln(\beta) - n\times\ln(\alpha+1) + (\alpha+1) \sum_1^n \ln(y_i)$
I then take the partials in terms of alpha and beta When I take it in terms of Beta I get $0=-\alpha n/\beta$ I do not know what to do with this if I can do anything at all
Similarly when I take the partials in terms of alpha I get a similar problem. I don't know what to do with it. Would appreciate it if someone could show me how to get to the solution.
As hinted in the comments, to compute the density, you have to differentiate the distribution function : $$f_i(y\mid\alpha,\beta) := \partial_y \mathbb P(Y_i \le y \mid \alpha,\beta) =\alpha \frac{ y^{\alpha-1}}{\beta^\alpha} \mathbf{1}_{0\le y \le \beta} $$ Now, the likelihood is given by the product of the conditional densities taken at each observation, and as is common, we'll rather consider the log-likelihood to convert the product into a sum : $$ \begin{align}\mathcal L(\alpha,\beta) &= \ln\left(\prod_i^n f_i(Y_i\mid\alpha,\beta)\right) \\ &= \ln\left(\prod_i^n \alpha \frac{ Y_i^{\alpha-1}}{\beta^\alpha} \mathbf{1}_{0\le Y_i \le \beta}\right) \\ &= n\ln\left(\frac{\alpha}{\beta^\alpha}\right) + (\alpha - 1)\sum_i^n \ln(Y_i \mathbf{1}_{0\le Y_i \le \beta}) \end{align} $$
The MLEs of $\alpha$ and $\beta$ are the values $\hat \alpha$ and $\hat \beta$ which maximize $\mathcal L(\alpha, \beta)$
For $\beta$ : We see that, for the likelihood not to be equal to $- \infty$, we necessarily need $\hat \beta \ge \max\{Y_1,\ldots,Y_n\}$. Since $\mathcal L(\alpha,\beta)$ is a non increasing function of $\beta$, we see that $$\hat \beta = \max\{Y_1,\ldots,Y_n\}$$ (the minimal acceptable value) is the MLE of $\beta$.
For $\alpha$ : $\partial_\alpha \mathcal L(\alpha, \beta) = \frac{n}{\alpha} - n\ln(\beta) + \sum \ln(Y_i \mathbf{1}_{0\le Y_i\le \beta})$
We find $$\hat \alpha = \frac{n}{\ln\left(\frac{{\hat \beta}^n}{\prod_i^n Y_i}\right)}$$