How Does the Constraint θ ≤ 1/4 Affect the MLE in Bernoulli Trials?

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Homework Statement


Two independent bernoulli trials resulted in one failure and one success. What is the MLE of the probability of success θ is it is know that θ is at most 1/4


Homework Equations


f(x,θ) = θx (1-θ)1-x


The Attempt at a Solution


Now, I know how to find the likelihood and use it to solve for the MLE. But I am not sure how the "θ is at most 1/4" would factor into the equation.

For a Bernoulli trial: f(x,θ) = θx (1-θ)1-x
L(θ) = θ(1-θ)
L'(θ) = 1-2θ ----> equate to zero
θ(hat) = 1/2 (which is the MLE)

But what do I do about the fact that "θ is at most 1/4"? Please help
 
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You're trying to maximize L(θ) on the interval [0,1/4]. Recall a basic calculus fact that the maximum of a function on a closed interval is either a critical point (where the derivative is zero) or an end point
 
Thank you.
In this case, since the function is simply: θ(hat) = 1/2 (a constant)
The maximum will lie at 1/4
Correct?
 
The critical point is at 1/2. But the function you want to maximize is L(θ) = θ(1-θ)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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