Arsthbt

I have to check that this propriety

$Z sim N(0,1)$ and $Usim chi ^2(10)$ then $ Z/sqrtU/10 sim T(10)$

is true using r studio if anyone can help , much appreciate

One approach could be simulation of thousands of values:

• Simulate $Z$ using rnorm


• Simulate $U$ using rchisq


• Do the division $Y = Z / sqrtU / 10$
  


• Simulate the same number of $T$ from the hypothesised $t$-distribution using rt


• Sort $Y$ and $T$ and plot them against each other - you want to see a diagonal straight line essentially $y=x$ with a little noise; this is visual demonstration though not a proof that the distributions are the same

You can do similar things with the qqplot function if you know what you are doing

I agree with @angryavian that you can't do a 'proof' in R.
Also, it is crucial to state that random variables $Z$
and $U$ are independent. Then $Y = fracZU/sqrt10 sim mathsfT(10)$ by definition.

Here is R code to simulate a million values of $T$ [as in the Answer of @Henry (+1)], then to compare their histogram with the density of $mathsfT(10).$ This
is a graphical demonstration that $T$ has (at least very nearly) the claimed t distribution.

set.seed(405) # for reproducibility
z = rnorm(10^6); u = rchisq(10^6, 10)
y = z/sqrt(u/10)
hist(y, prob=T, br=50, col="skyblue2")
curve(dt(x, 10), add=T, lwd=2)

Furthermore, you could check that the quantiles of $Y$ very nearly match the theoretical quantiles of $mathsfT(10).$

summary(y)
 Min. 1st Qu. Median Mean 3rd Qu. Max. 
-10.641101 -0.699409 0.000059 0.000221 0.701253 9.802922 
qt(c(.25,.5,.75), 10)
[1] -0.6998121 0.0000000 0.6998121

The summary above also shows that $bar Y approx 0.$ And the sample variance of the simulated values of $Y$ is very nearly the variance $nu/(nu - 2) = 10/8 = 1.25$ of Student's t distribution with $nu = 10$ degrees of freedom.
[In effect, two of the moments suggested by #GeorgeDewhirts (+1).]

var(y); 10/8
[1] 1.250115
[1] 1.25

Also, you could do a Kolmogorov-Smirnov goodness-of-fit test on the first 5000 values of $Y$ and check that the P-value exceeds 5%. (The K-S test in R is limited to 5000 observations.) Roughly speaking, this is a formal, quantitative way to do @Henry's comparison of sorted observations.

ks.test(y[1:5000], pt, 10)

 One-sample Kolmogorov-Smirnov test

data: y[1:5000]
D = 0.013661, p-value = 0.3083
alternative hypothesis: two-sided

You could compare the moments of your distribution with the theoretical moments of $T(10)$