Every substitution, Jacobian, boundary limit, Beta/Gamma identity, and mgf derivative made explicit. If it looks long… that’s the point.
Support & PDF: \(f(x)=\frac{1}{b-a}\,\mathbf 1_{[a,b]}(x)\), with \(a
Key idea: always compute variance as \(Var(X)=E[X^2]-(E[X])^2\).
PDF \(\phi_{\mu,\sigma}(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp\!\big(-\frac{(x-\mu)^2}{2\sigma^2}\big)\).
Show \(\int (x-\mu)\phi_{\mu,\sigma}(x)dx=0\) by oddness in the shifted variable; then compute \(E[(X-\mu)^2]\) directly.
PDF \(f(x)=\lambda e^{-\lambda x}\) on \([0,\infty)\).
If \(Y\sim\mathcal N(m,s^2)\), then \(M_Y(t)=\exp(mt+\tfrac{1}{2}s^2 t^2)\).
Let \(x=e^y\Rightarrow dx=e^y dy\), lognormal pdf \(f_X(x)=\dfrac{1}{x s\sqrt{2\pi}}\exp\!\left(-\dfrac{(\ln x - m)^2}{2s^2}\right)\). Then for \(r>-1\)
PDF: \(f(x)=\dfrac{x^{\alpha-1} e^{-x/\theta}}{\Gamma(\alpha)\,\theta^{\alpha}}\), \(x>0\).
Recall \(\Gamma(z)=\int_0^{\infty} t^{z-1}e^{-t}dt\). Then \(\Gamma(z+1)=z\Gamma(z)\) by integration by parts with \(u=t^{z}\), \(dv=e^{-t}dt\).
PDF on \((0,1)\): \(f(x)=\dfrac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}\), with \(B(\alpha,\beta)=\dfrac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\).
PDF: \(f(x)=\dfrac{k}{\lambda}(x/\lambda)^{k-1} e^{-(x/\lambda)^k}\), \(x>0\).
If \(X\sim\chi^2_{\nu}\), then \(X\sim\Gamma(\alpha=\nu/2,\theta=2)\).
If \(Z\sim N(0,1)\), then mgf of \(Z^2\) is \(M_{Z^2}(t)=(1-2t)^{-1/2}\) (compute \(E[e^{tZ^2}]\) by Gaussian integral). For independent \(Z_1,\dots,Z_{\nu}\), mgf of \(\sum Z_i^2\) is \((1-2t)^{-\nu/2}\). Differentiate: \(M'(t)=\nu(1-2t)^{-\nu/2-1}\), \(M''(t)=\nu(\nu+2)(1-2t)^{-\nu/2-2}\). At \(t=0\): \(E[X]=\nu\), \(E[X^2]=\nu(\nu+2)\), variance \(=2\nu\).