Shape of the distribution

The \(\ChiSqrDistn(1\;\text{df})\) distribution is very skew and its pdf is infinite at zero — \(f(0) = \infty\) — but whose right tail is very long.

Shape of the distribution

The diagram below initially shows the pdf of the absolute value of a standard normal variable, \(Z\). Its shape is that of the right half of the standard normal distribution.

Use the slider to apply a power transformation to \(Z\) — to display the pdf of \(Z^k\) for different powers \(k\). Drag the slider fully to the right to see the distribution of \(Z^2\) — a \(\ChiSqrDistn(1\;\text{df})\) distribution.

The transformation is nonlinear, so the shape of the pdf changes.

Click anywhere on the distribution to highlight some values of Z. Observe that the highlighted area (the probability) remains the same when you transform.

The mean and variance of the \(\ChiSqrDistn(1\;\text{df})\) distribution can be found directly from the following integrals.

\[ E[X] = \int_0^{\infty}x\;f(x) dx \spaced{and} E[X^2] = \int_0^{\infty}x^2\;f(x) dx \]

However since the distribution is a special case of the Gamma distribution, it is easier to refer to the earlier results about the mean and variance of the Gamma distribution.

Mean and variance

If a random variable \(X\) has a \(\ChiSqrDistn(1\;\text{df})\) distribution, its mean and variance are

\[ E[X] \;=\; 1 \spaced{and} \Var(X) \;=\; 2 \]

The mean and variance of a \(\GammaDistn(\alpha, \beta)\) distribution are

\[ E[X] \;=\; \frac{\alpha}{\beta} \spaced{and} \Var(X) \;=\; \frac{\alpha}{\beta^2} \]

Since the \(\ChiSqrDistn(1\;\text{df})\) distribution is identical to a \(\GammaDistn(\frac 1 2, \frac 1 2)\) distribution, its mean and variance can be found by substituting \(\alpha = \frac 1 2\) and \(\beta = \frac 1 2\).