Question

Statistics/Probability:

Suppose you are dealing with Normal and Chi-Square distributions.
Let \(X_i\sim\mathrm{N}(0,4), i=1,2,3.\) Let \(Y_j\sim\chi^2(4j-1), j=1,2.\) Find \(k\) for the following probability: \(P\left(X_1-X_2>k\sqrt{X_3^2+4(Y_1+Y_2)}\right) \) assuming all random variables are independent of each other.

Answer 1

Oops I realized I am missing something. The probability is P(...) = 0.95

Answer 2

\[P\left(X_1-X_2>k\sqrt{X_3^2+4(Y_1+Y_2)} \right)=P\left(\frac{X_1-X_2}{\sqrt{X_3^2+4

(Y_1+Y_2)}}>k \right) \]
I think you can make use of the t-distribution... Since the numerator will be normal, and

you have a square root of a sum of chi square random variables.

Now: For independent \(X_1, ... X_n \sim \text{N}(\mu_i, \sigma_i^2)\),
\(\large \sum\limits_{i=1}^na_iX_i\sim \text{N}\left( \sum\limits_{i=1}^na_i\mu_i, \sum

\limits_{i=i}^{n}a_i^2\sigma_i^2\right)\)

So: \(X=X_1-X_2\sim\text{N}\left((1)(0)+(-1)(0),(1)^2(4)+(-1)^2(4)\right) =\text{N}

(0,8)\)

\(Y_1\sim \chi^2(4(1)-1)=\chi^2(3)\)
\(Y_2\sim \chi^2(4(2)-1)=\chi^2(7)\)
In the denominator... you have a sum of Chi Squared r.v.s: \(W=Y_1+Y_2 \sim \chi^2(10)\)


So you have \[P\left( \frac{X}{\sqrt{X_3^2+4W}}>k\right)=0.95 \] To relate it to the t

distribution, you need to have \[\frac{Z}{\sqrt{S/\nu}} , \text{where } Z\sim N(0,1), ~S

\sim \chi^2(\nu) \] So by manipulating your random variables:
\[ P\left( \frac{\sqrt{8}\color{red}{\left(\frac{X}{\sqrt{8}}\right)}}{\sqrt{4\color

{blue}{\left (\frac{X_3}{2}\right)^2}+4W}}>k\right)\]
Where the part in red is a standard normal, and and the blue part is a standard normal

squared, so it is \(\chi^2(1)\) . I multiplied by the appropriate coefficient in front to

keep the expression equivalent. If you pull out the \(\sqrt{4}\), notice that you'll get a sum of 2 chi-square r.v.'s which will be \(\chi^2(11)\), so divide the top and bottom by \(\sqrt{11}\)
\[ P\left( \frac{\sqrt{8}\color{red}{\left(\frac{X}{\sqrt{8}}\right)}}{\frac{\color{orange}{\sqrt{11}}\cdot2\left

(\sqrt{\color{blue}{\left (\frac{X_3}{2}\right)^2}+W}\right)}{\color

{orange}{\sqrt{11}}}}>k\right)\] which hope you can identify the t-distribution in there. So you have \[P\left(\frac{\sqrt{8}}{2\sqrt{11}}T >k\right)=P\left(T>\frac{k\sqrt{11}}{\sqrt{2}} \right)=0.95\] where \(T\sim t(11)\) . Then you should find k easily now from a t-table or using a calculator/software.

Answer 3

I hope you can do that actually. Actually I don't remember that well if you can just pull out coefficients from random variables like I did when I factored out \(\sqrt{4}\). Maybe some second advice: @SithsAndGiggles

Answer 4

I think scalar multipliers generally follow the same rules as for the usual algebraic non-random variables, so I would think the work is valid. Nice job!