# Deriving Mean and Variance of (constant * Gaussian Random Variable) and (constant + Gaussian Random Variable)

If X is a Gaussian, N($$\mu_x, \sigma_x^2$$), then I came across the following properties:

1) Y = X + k, k is a constant. Then Y is normally distributed as N($$\mu_y, \sigma_y^2$$) such that:

$$\mu_y = \mu_x + k$$

$$\sigma_y = \sigma_x$$

2) Y = kX, k is a constant. hen Y is normally distributed as N($$\mu_y, \sigma_y^2$$) such that:

$$\mu_y = k*\mu_x$$

$$\sigma_y = |k|*\sigma_x$$

Is there a proof for the above two statements? Thanks in advance.

{ Edit: I have modified the original question.

Original question: Deriving Mean and Variance of (constant * Gaussian Distribution).

Due to some misunderstandings, I came to know that the above is not a feasible PDF and that what I actually want is: Deriving Mean and Variance of (constant * Gaussian Random Variable) and (constant + Gaussian Random Variable). }

Edit 2:

If X is Gaussian, prove kX is Gaussian where k is a constant such that k<0.

$$P(kX < x)$$

$$= P(X > \frac{x}{k})$$ (since k<0)

$$= \int_{\frac xk}^\infty \frac{1}{\sqrt{2\pi}\sigma}\exp \left(-\frac{(t-\mu)^2}{2\sigma^2}\right)dt$$

I am not able to resolve the limits to get the resulting equation into the form for Cumulative Density Function. How do I proceed?

• If $f(x)$ denotes a distribution's PDF, $cf(x)$ is only a PDF if $c=1$. Perhaps you meant $cf(cx)$. – J.G. Oct 21 '19 at 14:18
• Thank you. I see. No. I do mean c*f(x). So the definition of E(x) probably does not hold for this distribution. My reason for this question is, while learning Neural Nets I came across this idea that if f(x) is a standard normal distribution, multiplying it by square_root(2/n) where n is the number of nodes in the layer transforms the distribution to one with variance 2/n. I am looking for more intuition into this. Any help pointing me in the right direction appreciated. Will add this to the original question too. – Nitin Oct 21 '19 at 14:25
• It should be $\mu_y=\mu_x+k$ in the first one. – Milten Oct 23 '19 at 9:28
• And $\sigma_y=|k|\sigma_x$ in the second one, in case $k$ can be negative. – Milten Oct 23 '19 at 9:30
• Sorry. Yeah, I've updated it. – Nitin Oct 23 '19 at 9:43

The expressions for the mean and standard deviation are true for any random variable (EDIT: for which $$EX$$ and $$VX$$ are defined): $$E(X+k) = EX+Ek = EX + k, \quad V(X+k)=VX+Vk=VX,$$ $$E(kX) = kEX, \quad V(kX) = k^2VX$$ I am sure you can find proofs for these easily. But we still need to show that the transformed variables have normal distributions. We can do that as follows. Assume $$k>0$$. Then: $$P(kX $$= \int_{-\infty}^x \frac{1}{\sqrt{2\pi}k\sigma}\exp \left(-\frac{(u -k\mu)^2}{2(k\sigma)^2}\right)du$$ where we substituted $$u=kt$$. This is the CDF of $$\mathcal N(k\mu, k^2\sigma^2)$$, so $$kX$$ has that distribution. (Note that we rediscovered the mean and variance). The cases $$kX$$ for $$k<0$$, and $$X+k$$ can be shown similarly.
EDIT: The case of $$kX$$, $$k<0$$. You showed $$P(kX We substitute $$u=kt$$, so $$du=kdt$$. The lower limit becomes $$k\frac xk = x$$, and the upper limit becomes $$k\infty = -\infty$$ (you can rephrase this as a limit, if you want). Plugging in: $$\int_x^{-\infty} \frac 1k \cdot \frac{1}{\sqrt{2\pi}\sigma}\exp \left(-\frac{(\frac uk -\mu)^2}{2\sigma^2}\right)du$$ $$= -\int_x^{-\infty} \frac 1{|k|} \cdot \frac{1}{\sqrt{2\pi}\sigma}\exp \left(-\frac{(u -k\mu)^2}{2k^2\sigma^2}\right)du$$ $$= \int_{-\infty}^x \frac{1}{\sqrt{2\pi}|k|\sigma}\exp \left(-\frac{(u - k\mu)^2}{2(|k|\sigma)^2}\right)du$$ which is the CDF for $$\mathcal N(k\mu,(|k|\sigma)^2)$$.
• Thank you. Extending for $X + k$ is pretty simple. However, I am not able to extend for $kX$ when k<0. I have edited the question to add my workings. Could you help me please? – Nitin Oct 24 '19 at 16:21
• Correction: these columns of statements are true respectively if $E(X),\,V(X)$ are defined. – J.G. Oct 24 '19 at 16:29
• @Nitin You are close. You just need to make the substitution $u=kt$. I will edit my answer tomorrow to include it. – Milten Oct 24 '19 at 19:12