Writing proofs and solutions completely but concisely I have a university lecturer that puts a lot of emphasis on the writing quality of an answer, not just if its correct. He wants our answers to be 5 things, Clear, Complete, Concise, Coherent and Correct. He says he doesn't believe in perfect answers so can be very nit-picky over our solutions, but it also means he can be contradictory. For example in one of my solutions he said i had been repetitive but firstly, the repeated line (which was something like, "by the law of total variance") was there to help the narrative of the answer - which he puts so much emphasis on - and it was repeated from part 1 of a question after quite a few lines of integration. But secondly and more importantly, in his very own solution, he had the exact same repetition. And he said my answer was very good and he was being really nitpicky but that i could at some point write more detail, like write "the pdf of X is f(x)=..." rather than just go straight into "f(x)=..."
So my question is how do you strike a balance between explaining enough and becoming too laborious? Also what are some good connecting words in proofs and 'show me' questions and when do you use which? Because i just tend to stick in 'and', 'because', 'so' 'thus' etc, when prehaps sometimes it isn't strictly true (but does get the point across)
An example of such a question: Let X~Gamma($\alpha,\beta$) where $\beta$ is a rate parameter. Find the MGF of X and
use this to show that $\Bbb{E}(X)=\frac{\alpha}{\beta}$ and $var(X)=\frac{\alpha}{\beta^2}$
Let X~Gamma($\alpha , \beta$)
Then the PDF of X is:
$$f(x)=\frac{\beta^\alpha}{\Gamma(\alpha)}x^{(\alpha-1)}e^{-\beta x}dx $$ ,for $x>0$ and 0 otherwise, where $$\Gamma(\alpha)=\int_0^\inf t^{\alpha-1}e^{-t}dt$$
$$M_X(t)=\mathbb{E}(e^{tX})=\int_0^\inf e^{tx}f(x)dx$$
$$=\frac{\beta^\alpha}{\Gamma(\alpha)}\int_0^\inf e^{tx}x^{\alpha-1}e^{-\beta x}dx$$
$$=\frac{\beta^\alpha}{\Gamma(\alpha)}int_0^\inf x^{\alpha-1}e^{-(\beta-t) x}dx$$
Let $$u=(\beta-t)x$$
$$\iff x=\frac{u}{\beta-t}$$
$$\implies dx=(\beta-t)du$$
So $$M_X(t)=\frac{\beta^\alpha}{\Gamma(\alpha)}\int_0^\inf (\frac{u}{\beta-t})^{\alpha-1}e^-u(\beta-t)du$$
$$=\frac{\beta^\alpha}{\Gamma(\alpha)}(\frac{1}{\beta-t})^{\alpha}\int_0^\inf u^{\alpha-1}e^{-u}du$$
$$\frac{\beta^\alpha}{(\beta-t)^\alpha}$$
To show that $\Bbb{E}(X)=\frac{\alpha}{\beta}$ and $var(X)=\frac{\alpha}{\beta^2}$:
Using differentiation gives us:
$$M'(t)=\frac{\alpha\beta^\alpha}{(\beta-t)^{\alpha+1}}$$ and $$ M''(t)=\frac{\alpha(\alpha+1)\beta^\alpha}{(\beta-t)^{\alpha+21}}$$
It follows that
$$\Bbb{E}(X)=M'(0)=\frac{\alpha\beta^\alpha}{\beta^{\alpha+1}}=\frac{\alpha}{\beta}$$
and
$$\Bbb{E}(X^2)=M''(0)=\frac{\alpha(\alpha+1)\beta^\alpha}{\beta^{\alpha+21}}=\frac{\alpha(\alpha+1)}{\beta^2}$$
Hence
$$ var(X)=\Bbb{E}(X^2)-\Bbb{E}(X)^2$$
$$=\frac{\alpha(\alpha+1)}{\beta^2}-\frac{\alpha^2}{\beta^2}$$
$$=\frac{\alpha}{\beta^2}$$ as required
I know i need connecting words before i calculate the M(t) but i have no idea what to put, plus I'm not sure if where I've put "it follows that" is strictly true, also i feel like where i've used "So" there might be a better word to use but i can't think what
 A: Based on your lecturer's feedback, it seems like you are already doing a fairly good job! Most pieces of writing could be improved; maybe if you asked your lecturer about the "contradictory" remark, he'd say that his own answer could also be better.
If you're starting out writing proofs, it helps to have some rules of thumb to look at when you're writing. Many textbooks about proofs have a list; for example, here (section 5.3 on Mathematical Writing) is a list from Richard Hammack's Book of Proof.
The first point on that list explains why it is better to write "The PDF of $X$ is $f(x)=\dots$" rather than just "$f(x) = \dots$": beginning a sentence with a symbol is bad style (and sometimes confusing, because it doesn't look the way we expect a sentence to begin). If you didn't explain that $f(x)$ is the PDF of $X$, you could also be running afoul of point 8: always explain each new symbol you introduce.
Points 10 and 11 give you some guidance on how to use connecting words. If you want these to come more naturally to you, I suggest reading more proofs, and imitating the style of those that seem clearest to you.

Here is some concrete feedback on your writeup. Using words between equations is important, but the key phrase is using words; not just adding them. A single "so" doesn't add anything; you can and should say more. Two of the goals words can serve are:

*

*Separate different steps in your proof, so that it's not a surprise (for example) when you switch from talking about $\Gamma(\alpha)$ to talking about $M_X(t)$.

*Explain what you're about to do.

I would rewrite the first half of your proof as follows:

Then the PDF of $X$ is $$f(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}\,dx$$ for $x>0$ and $0$ otherwise, where $\Gamma(\alpha) = \int_0^\infty t^{\alpha-1} e^{-t}\,dt$.

(Some commentary: not every equation needs its own line. Here, it makes sense to put $\Gamma(\alpha)$ inline, since it's not as important: you're explaining something the reader should already know.)

Substituting this into the definition of $M_X(t) = \mathbb E(e^{t X})$, we get \begin{align} M_X(t) &= \int_0^\infty e^{tx} f(x)\,dx \\ &= \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty e^{tx} x^{\alpha-1}e^{-\beta x}\,dx \\ &= \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha-1} e^{-(\beta-t)x}\,dx.\end{align}

(I explain what we're going to be doing in this step: this is just substitution and some simplification. Note that it helps to align the equations to make it clear that all of this is just different expressions for $M_X(t)$.)

Now perform a $u$-substitution where $u = (\beta-t)x$ and, accordingly, $x = \frac{u}{\beta-t}$ and $dx = \frac1{\beta-t}\,du$. The result is \begin{align}M_X(t) &= \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty \left(\frac{u}{\beta-t}\right)^{\alpha-1} e^{-u} \frac1{\beta-t}\,du \\ &= \frac{\beta^\alpha}{\Gamma(\alpha)} \left(\frac{1}{\beta-t}\right)^\alpha \int_0^\infty u^{\alpha-1} e^{-u}\,du \\ &= \frac{\beta^\alpha}{(\beta-t)^{\alpha}}.\end{align}

(Again, equations like $u = (\beta-t)x$ aren't complicated enough to need their own line. On the other hand, explaining that we're going to do a $u$-substitution is worthwhile! Also, I corrected a mathematical error: $dx = (\beta-t)\,du$ should have been $dx = \frac1{\beta-t}\,du$. If I were to add anything else, I would add a sentence at the end explaining that we replace $\displaystyle \int_0^\infty u^{\alpha-1} e^{-u}\,du$ by $\Gamma(\alpha)$.)
