Brodudedoodebrodude

Method: Try to isolate the x from everything else

In this case:

1: Add x to both sides

2: Subtract 43 from both sides

3: Divide both sides by 4

You now have your answer (x= - 4)

If you're wondering how to factorise this, then you need to think of two numbers that:

-> Multiply together and give you -6

-> Add together to give you -1

Eventually you can do it in your head, but for now it's probably easiest to write down all the whole numbers that multiply together to give you -6;

-1x6 = -6    1x-6 = -6

-2x3 = -6    2x-3 = -6

Now from these numbers which ones add together to give you negative one?

-3 + 2 = -1

So the two numbers you needed are 2 and -3.

Therefore it factorises into \(z^2-z-6 = (z+2)(z-3)\)

Remember that when you divide or multiply by a negative then you have to switch the sign of the inequality:

\(2x+4 < 5x + 3 \\ 2x+4 -4 < 5x+3 - 4 \\ 2x < 5x-1 \\ 2x-5x<5x-1-5x\\-3x<-1\\ \frac{-3x}{-3}>\frac{-1}{-3}\\x>\frac{1}{3}\)

Kind of late reply as I havn't been on but that's what I ended up doing so your answer agrees with me -> The M.C consists of 4 states (0,0) , (1,0) , (0,1) , (1,1) and then you find the transition probabilities for each state using what you've given.

At the bottom of the summation it indicates what it starts on, in this case i starts on 1. At the top of the summation it tells you what the maximum number is that i gets up to. Now you need to sum every number from what you start on up till the maximum.

Your 1st sum translates to $$x_1+x_2+x_3+x_4+x_5+x_6=15$$Your 2nd sum translates to $$x_1+x_2+x_3+x_4+x_5=20$$Your 3rd sum translates to $$\frac{x_1+x_2+x_3+x_4+x_5+x_6+x_7}{7}=8$$

Hope you should be able to do the rest of the problem now =)

The 'tricks' for the first one is that you can take a constant outside of an integral and also that you can split up a single integral of many terms into many integrals of single terms:

$$E[ag(X)+bh(X)]= \int_{-\infty}^{\infty} [ag(X)+bh(X)]f_x(x)dx \\\indent= \int_{-\infty}^{\infty}ag(X)f_x(x)dx+\int_{-\infty}^{\infty}bh(X)f_x(x)dx \\\indent =a\int_{-\infty}^{\infty}g(X)f_x(x)dx+b\int_{-\infty}^{\infty}h(X)f_x(x)dx \\\indent = aE[g(X)]+bE[h(x)]$$

Part b is a very similar process to part a:

$$Var(aX) = E[(aX)^2]-(E[aX])^2 \\\indent =\int_{-\infty}^{\infty}a^2X^2f_x(x)dx-(\int_{-\infty}^{\infty}aXf_x(x)dx)^2\\\indent= a^2\int_{-\infty}^{\infty}X^2f_x(x)dx-(a\int_{-\infty}^{\infty}Xf_x(x)dx)^2 \\\indent= a^2E[X^2]-a^2(\int_{-\infty}^{\infty}Xf_x(x)dx)^2 \\\indent= a^2(E[X^2]-(E[X])^2)\\\indent=a^2Var(X)$$

I'm not too sure with part c so hopefully someone else can help with that =)

I understand it now, thanks a lot Alan =)

Yeah it does thanks Alan. What's still causing some confusion for me is what is the point of mentioning that the initial distribution for the markov chain is (1/3 , 1/3, 1/3)?

These are the answers provided by my lecturer but yes I also think it's a typo and should be -3 rather than 3, forgot to mention that! 

My problem is just the fundamentals, how does the calculation for P(0.1) look like? Once I see it I can make sense of it (I hope)

A verticle asymptote occurs where a value of x makes the denominator equal to zero but not the numerator:

In this case the only value of x that makes the denominator = 0 is x=-1/2 and this value also doesn't make the numerator = 0 therefore the vertical asymptote is at x=-1/2.

What if the function was this however: $$y(x)=\frac{(x^2+7x+10)(2x+1)}{2x+1}$$

If you dont cancel the 2x+1 and substitute in x=-1/2 you will get 0 for both the numerator and the denominator therefore this wouldnt be  a vertical asymptote. The way to go about this situation is to cancel the 2x+1 to get $$y(x) = x^2+7x+10$$ but you Have to mention that x cannot equal -1/2 as this isn't defined for the original function and is a discontinuity and then you can leave it at that.

Saul from Khan Academy Explains this nicely: https://www.khanacademy.org/math/algebra2/rational-expressions/rational-function-graphing/v/finding-asymptotes-example