# How to simulate sum of two normal distributed variables when linearly transformed?

I m trying to replicate a paper's simulation results, but am not having much success. The equation is: $$P_t = \beta_{PD}*X_t + \varepsilon_{PD,t}$$

• $P_t$ (by assumption) is described as standard normal.
• $\beta_{PD}$ is a constant $0\leq\beta_{PD}\leq1$.
• $\varepsilon_{PD,t}$ - is a residual term.

More precisely, $\varepsilon_{PD,t}$ are described in the paper as being "..residual changes, mutually independent, independent of $X_t$, and are normally distributed such that $P_t$ is a standard normal distributed variable", page 5. The sum of the variances of the transformed random variables should be $1$ as $P_t$ is $\mathcal N(0,1)$. The variance of the first term is $\beta_{PD}^2$/standard deviation is $\beta_{PD}$. How do you construct the second term such that the sum of the terms is $\mathcal N(0,1)$? Any other suggestions other than how I have done this below?

R code:

Bpd<-0.4; Xt<-rnorm(100000,0,1);E.pd<-rnorm(100000,0,1)
# eq 1 as stated in the paper
Pt = (Bpd)*Xt + E.pd
# makes Pt2 Standard N(0,1) but not as written (eq1) in the paper
Pt2<-(Bpd)^0.5*Xt+(1-Bpd)^0.5*E.pd
sd(Pt); sd(Pt2)


Reference: Assessing the Relationship between Probability of Default and Loss Given Default in an Agricultural Loan Portfolio, by Nicholas K. Sakaimbo and Glenn D. Pederson.

-

This isn't possible unless the mean of epsilon depends on X (or$\beta=0$) and I think it most likely there is a mistake in the paper.

X is described as the systemic risk factor and I think is meant to be non random so that simplifies things somewhat.

The mean of $\varepsilon_{PD,t}$ must vary depending on PD and t, as it must exactly cancel out the structural part of the model (ie $E(\varepsilon_{PD,t})=-\beta_{PD}*X_t$. This seems so odd that I can't imagine it is what they mean, although it does literally meet their definition.

If you want to simulate this in R rnorm() can take a vector of variable means to create the epsilons (with variance one and mean defined as above); but I doubt this would be worthwhile.

> Bpd<-0.4; Xt<-rnorm(100000,0,1)
>
> E.pd<-rnorm(100000,-Bpd*Xt,1)
>
> Pt = (Bpd)*Xt + E.pd
> c(mean(Pt), var(Pt))
[1] 0.004140578 0.997591190

-
oops, just spotted the definition says the epsilons are independent of X. So it simply isn't possible as described in the paper. –  Peter Ellis Jan 14 '13 at 19:09
Thanks for the reply Peter (@peter). Lets forget about the t subscripts for clarity. X is standard normal random variable, it is multiplied by Beta, a constant. This product plus the mysterious epsilon term should sum to a standard normal variable. As the paper is pretty vague, I think resorting to authors is next. In any event, thank you for the reply. –  WoodsT Jan 15 '13 at 9:58