Find the maximumlikelihood estimator of the following density function

Question

Find the maximumlikelihood estimator of the following density function.

When I asked before, @Glen_b said that this is special case of shifted gamma distribution.

$$y=a+u^2$$ for u~N(0, $\sigma^2$)

Question says that work out the density of y.

Therefore I take

$f(y,a)=(y-a)^{1/2}$

At this point I am not sure whether I use such a function or not. What is the density function of y?

But I assumed that I should use this density function of y. And I calculate as follows:

Firstly I write the joint density

$L(a)=\prod_{i=1}^n (y_i-a)^{1/2}$

And then I take its logarithm,

$ln(L(a))=ln[\prod_{i=1}^n (y_i-a)^{1/2}]={1\over 2}\sum_{i=1}^nln(y_i-a)$

Finally, take its derivative with respect to a,

${\partial ln(L(a))\over \partial a}={-1\over 2}\sum_{i=1}^n{1\over (y_i-a)}=0$

But I cannot derive $a$ alone to find $\hat{a}_{ML}$

I guess my solution have some mistakes. Especially I think $f(y,a) $ is false. But I don't know how I can express it.

Please show me something /my mistakes about it. If I write $f(y,a) $ correctly, I can find $\hat{a}_{ML}$ by myself.

Thank you in advance by helps.

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moved up from comments:

$u=\sqrt{x-a}$ and $u=-\sqrt{x-a}$ $|du/dy|=|1/2 (y-a)^{-1/2}|$ And I use this formula $f_y=|du/dy|f(u)$ so my attempted answer for the density of $y$ is:

${1\over 2 \sigma \sqrt {2\pi ({y-a})}}exp(-(y-a)/2\sigma^2 ) $

Answer 1

Your title mentions MLE but you actually ask about calculating the CDF of $Y$ (which is what I address below).

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Consider this problem:

Let $U\sim \text{Unif}(-3,3)$, let $V=|U|$. Write down the density of $V$.

We can do this by inspection.

$f_U(u) = \frac16 \,,\quad -3<u<3$ and $0$ elsewhere.

Since all the density below $0$ will be taken over to the corresponding value the other side of $0$ (-2 maps to 2 and so forth), the density will be everywhere doubled:

$f_V(v) = \frac13 \,,\quad 0\leq v<3$ and $0$ elsewhere.

Or by basic probability reasoning:

Note that $F_U(u) = \begin{cases} \quad 0,\qquad\qquad u\leq -3,\\ (u+3)/6,\: -3<u<3,\\ \quad 1,\qquad\qquad u\geq 3 \end{cases}$

$F_V(v)=P(V\leq v) = P(|U|\leq v) = P(-v\leq U\leq v)=P(U\leq v)-P(U< -v)$

For $v$ between $0$ and $3$ that's $(v+3)/6 - (-v+3)/6 = 2v/6 = v/3$. Checking the remaining cases, we can write

$F_V(v) = \begin{cases} \quad 0,\qquad\qquad v< 0,\\ v/3,\qquad\: 0\leq v<3,\\ \quad 1,\qquad\qquad v\geq 3 \end{cases}$

We can get the densiy of $V$ from there by differentiation.

You should be able to do the $N(0,\sigma^2)$ case similarly to either the first or the second approach there.

Now consider that we want the density of $W=V^2$.

$F_W(w)=P(W\leq w) = P(V^2 \leq w) $ (note that this is monotonic in $V$ when $V$ is nonnegative, so we can just take the $\sqrt{}$ of both sides of the inequality)

$\qquad\quad\, = P(V\leq \sqrt{w}) = F_V(\sqrt{w})$ ...

.. and just focusing at the case where there's some density for now:

$\qquad\: =\sqrt{w}/3,\: 0\leq w<9$

hence by differentiation the density in that region is $\frac{1}{6\sqrt{w}}$ (and looking at the remaining cases is $0$ elsewhere)

Again you can do the normal case similarly.

If you can follow all that, you might try this one: $W_2 = W+2$. Again this can either be done by inspection or by basic probability manipulations as before.

You should then be able to put these together to write down the density of $|u|$, of $u^2$ and hence of $u^2+a$ in your original problem.

Once you get used to such manipulations you can do it all in one step. Note that there's a formula for the density with monotonic transformations but I suggest you get comfortable with reasoning it out as above before you go near the formula (once you know how to do the above, you can derive the formula any time you want it using an argument like we did here).

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I'll outline the approach I'd suggest you take here.

1. $U\sim N(0,\sigma^2)$ -- write down the density for $U$

2. Let $V=|U|$ -- write down the density for $V$.

   (you should get $\frac{2}{\sqrt{2\pi}\,\sigma}e^{-\frac{v^2}{2\sigma^2}},\:v>0$ - let's not bother to simplify it because that 2 out the front is going away in a minute)

3. Let $W=V^2$ ... this is now monotonic, which makes your life easier. Derive the density of $W$ (a scaled chi-squared 1) in a similar way to the above However in this case you never wrote $F_V(v)$ down, so you'll just end up with $F_V(\sqrt{w})$. That's okay because we don't need to write down $F_V$. When you take its derivative, you get $f_V(\sqrt{w}).\frac{d}{dw}\sqrt{w}$, which you can write straight down. The first term is just substitute $\sqrt{w}$ everywhere the density for $v$ has a $v$ and the second term (the Jacobian) is simple.

4. Let $Y=W+a$. Again, you should be able to do that step by step

If you can do all the steps one at a time, you might try dealing with it in a single transformation but you have to be very careful about how the densities relate, because the transformation isn't invertible.

The issues you had were all algebraic rather than probability. Nevertheless, you may want a basic book on mathematical statistics to practice such manipulations in that context. Perhaps Mendenhall et al (Mathematical Statistics with Applications) -- doesn't need to be a recent edition - though there are many others that would do about as well.

Note that there are numerous handy posts on transformations done here on site. See here & here & here, and here. That first one there gives the formula for the density of a monotonic transformation: $f_Y(y) = \frac{d}{dy} F_X(t^{-1}(y)) = f_X(t^{-1}(y))\cdot |\frac{d}{dy}t^{-1}(y)|$. Oh, this one may also be quite helpful